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5910271583103986a1b6749ec6d74d50 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | Let $K$ be the intersection of $BC$ and $AE$ . Since the radii of the two circles are 1:4, so we have $AD:AE=1:4$ , and the distance from $B$ to line $l$ and the distance from $C$ to line $l$ are in a ratio of 4:1, so $BK:CK=4:1$ . We can easily calculate the length of $BC$ to be 4, so $CK=\frac{4}{3}$ . Let $J$ be the foot of perpendicular line from $A$ to $BC$ , we can know that $BJ:CJ=1:4$ , so $BJ = 0.8$ $CJ=3.2$ $AJ=1.6$ , and $AK=\sqrt{1.6^2+\left(3.2+\frac{4}{3}\right)^2}=\frac{4}{3}\sqrt{13}$ . Since $CK^2 = EK\cdot AK$ , so $EK=\frac{4}{39}\sqrt{13}$ , and $AE = \frac{4}{3}\sqrt{13} - \frac{4}{39}\sqrt{13}= \frac{16}{13}\sqrt{13}$ $\sin\angle AKB=\frac{AJ}{AK} = \frac{1.6}{\frac{4}{3}\sqrt{13}}=\frac{1.2}{\sqrt{13}}$ , so the distance from $C$ to line $l$ is $d=CK\cdot \sin\angle AKB = \frac{4}{3}\cdot \frac{1.2}{\sqrt{13}}=\frac{1.6}{\sqrt{13}}$ . so the area is \[[ACE] = \frac{1}{2}\cdot AE\cdot d = \frac{1}{2}\cdot\frac{16}{13}\sqrt{13}\frac{1.6}{\sqrt{13}} = \frac{64}{65}\] The final answer is $\boxed{129}$ | null | 129 |
5910271583103986a1b6749ec6d74d50 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | Consider the common tangent from $A$ to both circles. Let this intersect $BC$ at point $K$ . From equal tangents, we have $BK=AK=CK$ , which implies that $\angle BAC = 90^\circ$
Let the center of $\mathcal{P}$ be $O_1$ , and the center of $\mathcal{Q}$ be $O_2$ . Angle chasing, we find that $\triangle O_1DA \sim \triangle O_2EA$ with a ratio of $1:4$ . Hence $4AD = AE$
We can easily deduce that $BC=4$ by dropping an altitude from $O_1$ to $O_2C$ . Let $\angle ABC = \theta$ . By some simple angle chasing, we obtain that $\angle BO_1A = 2\angle BDA = 2\angle ABC = 2\theta,$ and similarly $\angle CO_2A = 180 - 2\theta$
Using LoC, we get that $AB = \sqrt{2-2\cos2\theta}$ and $AC = \sqrt{32+32\cos2\theta}$ . From Pythagorean theorem, we have \[AB^2 + AC^2 = BC^2 \implies \cos 2\theta = -\frac{3}{5} \implies \cos \theta = \frac{1}{\sqrt5}, \sin \theta = \frac{2}{\sqrt 5}\] In other words, $AB = \frac{4}{\sqrt5}, AC = \frac{8}{\sqrt5}$
Using the area condition, we have: \begin{align*} \frac12 AD*AB \sin \angle DAB &= \frac 12 AE*AC \sin(90-\angle DAB) \\ AD*\frac{4}{\sqrt5} \sin \angle DAB &= 4AD*\frac{8}{\sqrt5} \cos \angle DAB \\ \sin \angle DAB &= 8 \cos \angle DAB \\ \implies \sin \angle DAB &= \frac{8}{\sqrt{65}} \end{align*}
Now, for brevity, let $\angle D = \angle ADB$ and $\angle A = \angle DAB$
From Law of Sines on $\triangle ABD$ , we have \begin{align*} \frac{AB}{\sin \angle D} &= \frac{AD}{\sin (180-\angle A - \angle D)} \\ \frac{\frac{4}{\sqrt5}}{\frac{2}{\sqrt5}} &= \frac{AD}{\sin \angle A \cos\angle D + \sin \angle D\cos\angle A} \\ 2 &= \frac{AD}{\frac{2}{\sqrt{13}}} \\ AD &= \frac{4}{\sqrt{13}} \end{align*}
It remains to find the area of $\triangle ABD$ . This is just \[\frac12 AD*AB*\sin \angle A = \frac12 * \frac{4}{\sqrt{13}}*\frac{4}{\sqrt5}*\frac{8}{\sqrt{65}} = \frac{64}{65}\] for an answer of $\boxed{129}.$ | null | 129 |
5910271583103986a1b6749ec6d74d50 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | Add in the line $k$ as the internal tangent between the two circles. Let $M$ be the midpoint of $BC$ ; It is well-known that $M$ is on $k$ and because $k$ is the radical axis of the two circles, $AM=BM=CM$ . Therefore because $M$ is the circumcenter of $\triangle{BAC}$ $\angle{BAC}=90^{\circ}$ . Let $O_P$ be the center of circle $\mathcal{P}$ and likewise let $O_Q$ be the center of circle $\mathcal{Q}$ . It is well known that by homothety $O_P, A,$ and $O_Q$ are collinear. It is well-known that $\angle{ABC}=\angle{ADB}=b$ , and likewise $\angle{ACB}=\angle{AEC}=c$ . By homothety, $AD=4AE$ , therefore since the two triangles mentioned in the problem, the length of the altitude from $B$ to $AD$ is four times the length of the altitude from $C$ to $AE$ . Using the Pythagorean Theorem, $BC = 4$ . By angle-chasing, $O_{P}AMB$ is cyclic, and likewise $O_{Q}CMA$ is cyclic. Use the Pythagorean Theorem for $\triangle{O_{P}BM}$ to get $O_{P}M=\sqrt{5}$ . Then by Ptolemy's Theorem $AB=\frac{4\sqrt{5}}{5} \implies AC=\frac{8\sqrt{5}}{5}$ . Now to compute the area, using what we know about the length of the altitude from $B$ to $AD$ is four times the length of the altitude from $C$ to $AE$ , letting $x$ be the length of the altitude from $C$ to $AE$ $x=\frac{8}{5\sqrt{13}}$ . From the Law of Sines, $\frac{BD}{\sin{A}}=2 \implies \sin{A}=\frac{4x}{AB} \implies BD=\frac{8x}{AB}=\frac{16\sqrt{65}}{65}$ . Then use the Pythagorean Theorem twice and add up the lengths to get $AD=\frac{4}{\sqrt{13}}$ . Use the formula $\frac{b \times h}{2}$ to get $\frac{64}{65} = \boxed{129}$ as the answer. | null | 129 |
5910271583103986a1b6749ec6d74d50 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | Let $P$ and $Q$ be the centers of circles $\mathcal{P}$ and $\mathcal{Q}$ , respectively.
Let $M$ be midpoint $BC, \beta = \angle ACB.$
Upper diagram shows that
$\sin 2\beta = \frac {4}{5}$ and $AC = 2 AB.$ Therefore $\cos 2\beta = \frac {3}{5}.$
Let $CH\perp l, BH'\perp l.$ Lower diagram shows that
$\angle CAE = \angle ABH' = \alpha$ (perpendicular sides)
and $\angle CQE = 2\alpha$ (the same intersept $\overset{\Large\frown} {CE}).$ \[\tan\alpha = \frac {1}{8}, \sin2\alpha = \frac{2 \tan \alpha}{1 + \tan^2 \alpha} = \frac {16}{65}, \cos2\alpha = \frac{1 - \tan^2 \alpha}{1 + \tan^2 \alpha} = \frac {63}{65}.\] The area \[[ACE] = [AQC]+[CQE]– [AQE].\] Hence \[[ACE] =\frac{AQ^2}{2} \left(\sin 2\alpha + \sin 2\beta - \sin(2\alpha + 2\beta)\right),\] \[[ACE] = 8\left( \frac{16}{65}+\frac{4}{5} - \frac{4}{5}\cdot \frac{63}{65} - \frac{3}{5}\cdot \frac{16}{65}\right) = \frac{64}{65}\implies \boxed{129}.\] vladimir.shelomovskii@gmail.com, vvsss | null | 129 |
5910271583103986a1b6749ec6d74d50 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | We begin by extending $\overline{AB}$ upwards until it intersects Circle $\mathcal{Q}$ . We can call this point of intersection $F$ . Connect $F$ with $E$ $C$ , and $A$ for future use.
Create a trapezoid with points $B$ $C$ , and the origins of Circles $\mathcal{P}$ and $\mathcal{Q}$ . After quick inspection, we can conclude that the distance between the origins is 5 and that $\overline{BC}$ is 4.
(Note: It is important to understand that we could simply use the formula for the distance between the tangent points on a line and two different circles, which is $2\cdot \sqrt{a\cdot b}$ , where $a$ and $b$ are the two respective radii of the circles. In our case, we get $2\cdot \sqrt{4} = 4$
Using similar triangles or homotheties, $AE=4\cdot AD$ and $AF=4\cdot BA$ $BC^2 = BA\cdot BF$ \[16 = BA\cdot (5\cdot BA)\] \[BA = \frac{4}{\sqrt{5}}\] \[AF = \frac{16}{\sqrt{5}}\] \[BF = 4\cdot \sqrt{5}\]
Inspecting $\triangle{BFC}$ , we recognize that it is a right triangle ( $\angle{BCF} = 90$ ) as the final length ( $\overline{FC}$ ) being 8 would allow for an $x-2x-x\sqrt{5}$ triangle. Hence, the diameter of circle $\mathcal{Q}$ $CF$ . This also means that $\angle{CAF} = \angle{CEF} = 90$
From the fact that $\triangle{ABC}$ is a right triangle: \[AB^2 + AC^2 = BC^2\] \[AC = \frac{8}{\sqrt{5}}\] (Note: We could have also used $\triangle{FAC}$ .)
Our next step is to start angle chasing to find any other similar triangles or shared angles. Label the intersection between $\overline{FC}$ and $\overline{DE}$ as $M$ . Label $\angle{CAE} = \angle{CFE} = \theta$ . Since $\angle{BAC} = 90$ $\angle{EAF} = \angle{DAB} = 90-\theta$ . Now, use the sine formula and the fact that the areas of $\triangle{DBA}$ and $\triangle{ACE}$ are equal to get: \[\frac{1}{2}\cdot AC\cdot AE\cdot \sin{\theta} = \frac{1}{2}\cdot AD\cdot AB\cdot \sin{(90-\theta)}\] Since $\sin{(90-\theta)} = \cos{\theta}$ \[\tan{\theta} = \frac{AD}{2\cdot AE}\]
Using right $\triangle{FCE}$ , since $\angle{ACM} = \angle{CFE}$ $\tan{\theta} = \frac{CE}{FE}$ . Hence, plugging into the previous equation: \[\frac{CE}{FE} = \frac{AD}{2\cdot AE}\] Using the Pythagorean theorem on $\triangle{FCE}$ $FE = \sqrt{(64 - CE^2)}$ . We also know that $AE=4\cdot AD$ Plugging back in: \[\frac{CE}{\sqrt{(64 - CE^2)}} = \frac{AD}{2\cdot (4\cdot AD)}\] \[\frac{CE}{\sqrt{(64 - CE^2)}} = \frac{1}{8}\] .
From here, we can square both sides and bring everything to one side to get: \[EF^2 + 64EF - 64 = 0\] \[EF = \frac{64}{\sqrt{65}}\] \[CE = \frac{8}{\sqrt{65}}\] We should also return to the fact that $\sin{\theta} = \frac{CE}{CF}$ from $\triangle{FCE}$ , so \[\sin{\theta} = \frac{1}{\sqrt{65}}\]
From the fact that $\angle{CAF} = \angle{CEF} = 90$ , we can use Ptolemy's Theorem on quadrilateral $ACEF$ $AC\cdot EF + CE\cdot FA = CF\cdot AE$ . Plugging in and solving, we get that $AE = \frac{16}{\sqrt{13}}$
We now have all of our pieces to use the Sine Formula on $\triangle{ACE}$ \[\frac{1}{2}\cdot AC\cdot AF\cdot \sin{\theta}\] \[\frac{1}{2}\cdot \frac{8}{\sqrt{5}}\cdot \frac{16}{\sqrt{13}}\cdot \frac{1}{\sqrt{65}} = \frac{64}{65} = \boxed{129}\] | null | 129 |
5910271583103986a1b6749ec6d74d50 | https://artofproblemsolving.com/wiki/index.php/2015_AIME_II_Problems/Problem_15 | Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$ , respectively, and are externally tangent at point $A$ . Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$ . Points $B$ and $C$ lie on the same side of $\ell$ , and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
[asy] import cse5; pathpen=black; pointpen=black; size(6cm); pair E = IP(L((-.2476,1.9689),(0.8,1.6),-3,5.5),CR((4,4),4)), D = (-.2476,1.9689); filldraw(D--(0.8,1.6)--(0,0)--cycle,gray(0.7)); filldraw(E--(0.8,1.6)--(4,0)--cycle,gray(0.7)); D(CR((0,1),1)); D(CR((4,4),4,150,390)); D(L(MP("D",D(D),N),MP("A",D((0.8,1.6)),NE),1,5.5)); D((-1.2,0)--MP("B",D((0,0)),S)--MP("C",D((4,0)),S)--(8,0)); D(MP("E",E,N)); [/asy] | Let the center of the larger circle be $O,$ and the center of the smaller circle be $P.$ It is not hard to find the areas of $ACO$ and $ABP$ using pythagorean theorem, which are $\frac{32}{5}$ and $\frac{2}{5}$ respectively. Assign $\angle AOC=a,\angle COE=c,\angle DPB=d,\angle BPA=b.$ We can figure out that $\angle AOE=\angle DOA=\theta$ using vertical angles and isosceles triangles. Now, using $[ABC]=\frac{1}{2}ab\sin C$ \[[ACE]=[AOC]+[COE]-[AOC]=\dfrac{32}{5}-8\sin c-8\sin \theta,\] \[[DAB]=[APD]+[APB]+[BPD]=\dfrac{2}{5}+\dfrac{1}{2}\sin \theta+\dfrac{1}{2}\sin d.\] We can also figure out that $\sin a=\dfrac{4}{5},\cos a=\dfrac{3}{5},\sin b=\dfrac{4}{5}, \cos b=-\dfrac{3}{5}.$ Also, $c=\theta-a$ and $d=360-\theta-b.$ Using sum and difference identities: \[\sin c=\dfrac{3}{5}\sin \theta-\dfrac{4}{5}\cos \theta,\] \[\sin d=\dfrac{3}{5}\sin\theta-\dfrac{4}{5}\cos\theta.\] (We can also notice that $c+d=360-\theta-b+\theta-a=360-(a+b)=180$ which means that $\sin c=\sin d.$ )
Substituting in the equations for $\sin c$ and $\sin d$ into the equations for $[ACE]$ and $[DAB],$ setting them equal, and simplifying: \[3=2\sin\theta+3\cos\theta.\] Solving this equation we get that $\sin\theta=\frac{12}{13}$ and $\cos\theta=\frac{5}{13}.$ Doing a lot of substitution gives us \[[ACE]=[DAB]=\dfrac{64}{65},\] which means the answer is $64+65=\boxed{129}.$ | null | 129 |
15635440fc152d7be55bae38089fa152 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_1 | The 8 eyelets for the lace of a sneaker all lie on a rectangle, four equally spaced on each of the longer sides. The rectangle has a width of 50 mm and a length of 80 mm. There is one eyelet at each vertex of the rectangle. The lace itself must pass between the vertex eyelets along a width side of the rectangle and then crisscross between successive eyelets until it reaches the two eyelets at the other width side of the rectangle as shown. After passing through these final eyelets, each of the ends of the lace must extend at least 200 mm farther to allow a knot to be tied. Find the minimum length of the lace in millimeters.
[asy] size(200); defaultpen(linewidth(0.7)); path laceL=(-20,-30)..tension 0.75 ..(-90,-135)..(-102,-147)..(-152,-150)..tension 2 ..(-155,-140)..(-135,-40)..(-50,-4)..tension 0.8 ..origin; path laceR=reflect((75,0),(75,-240))*laceL; draw(origin--(0,-240)--(150,-240)--(150,0)--cycle,gray); for(int i=0;i<=3;i=i+1) { path circ1=circle((0,-80*i),5),circ2=circle((150,-80*i),5); unfill(circ1); draw(circ1); unfill(circ2); draw(circ2); } draw(laceL--(150,-80)--(0,-160)--(150,-240)--(0,-240)--(150,-160)--(0,-80)--(150,0)^^laceR,linewidth(1));[/asy] | The rectangle is divided into three smaller rectangles with a width of 50 mm and a length of $\dfrac{80}{3}$ mm. According to the Pythagorean Theorem (or by noticing the 8-15-17 Pythagorean triple), the diagonal of the rectangle is $\sqrt{50^2+\left(\frac{80}{3}\right)^2}=\frac{170}{3}$ mm. Since that on the lace, there are 6 of these diagonals, a width, and an extension of at least 200 mm on each side. Therefore, the minimum of the lace in millimeters is \[6\times \dfrac{170}{3}+50+200\times 2=\boxed{790}.\] | null | 790 |
d98a450075ec9a2da9e29396bc1b9fd7 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_2 | An urn contains $4$ green balls and $6$ blue balls. A second urn contains $16$ green balls and $N$ blue balls. A single ball is drawn at random from each urn. The probability that both balls are of the same color is $0.58$ . Find $N$ | First, we find the probability both are green, then the probability both are blue, and add the two probabilities. The sum should be equal to $0.58$
The probability both are green is $\frac{4}{10}\cdot\frac{16}{16+N}$ , and the probability both are blue is $\frac{6}{10}\cdot\frac{N}{16+N}$ , so \[\frac{4}{10}\cdot\frac{16}{16+N}+\frac{6}{10}\cdot\frac{N}{16+N}=\frac{29}{50}\] Solving this equation, \[20\left(\frac{16}{16+N}\right)+30\left(\frac{N}{16+N}\right)=29\] Multiplying both sides by $16+N$ , we get
\begin{align*} 20\cdot16+30\cdot N&=29(16+N)\\ 320+30N&=464+29N\\ N&=\boxed{144} | null | 144 |
92fc8a80dff5911d9be553ea46895107 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | Let the numerator and denominator $x,y$ with $\gcd(x,y)=1$ and $x+y = 1000.$ Now if $\gcd(x,y) = 1$ then $\gcd(x,y) =\gcd(x,1000-x)= \gcd(x,1000-x-(-1)x)=\gcd(x,1000)=1.$ Therefore any pair that works satisfies $\gcd(x,1000)= 1.$ By Euler's totient theorem, there are $\phi(1000) = 400$ numbers relatively prime to 1000 from 1 to 1000. Recall that $r=\frac{x}{y}<1$ and note by Euclidean algorithm $\gcd(1000,1000-x)=1$ , so we want $x<y=1000-x.$ Thus the $400$ relatively prime numbers can generate $\boxed{200}$ desired fractions. | null | 200 |
92fc8a80dff5911d9be553ea46895107 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | If the initial manipulation is not obvious, consider the Euclidean Algorithm. Instead of using $\frac{n}{m}$ as the fraction to use the Euclidean Algorithm on, we can rewrite this as $\frac{500-x}{500+x}$ or \[\gcd(500+x,500-x)=\gcd((500+x)+(500-x),500-x)=\gcd(1000,500-x).\] Thus, we want $\gcd(1000,500-x)=1$ . You can either proceed as Solution $1$ , or consider that no even numbers work, limiting us to $250$ choices of numbers and restricting $x$ to be odd. If $x$ is odd, $500-x$ is odd, so the only possible common factors $1000$ and $500-x$ can share are multiples of $5$ . Thus, we want to avoid these. There are $50$ odd multiples of $5$ less than $500$ , so the answer is $250-50=\boxed{200}$ | null | 200 |
92fc8a80dff5911d9be553ea46895107 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | Say $r=\frac{d}{1000-d}$ ; then $1\leq d\leq499$ . If this fraction is reducible, then the modulus of some number for $d$ is the same as the modulus for $1000-d$ . Since $1000=2^3\cdot5^3$ , that modulus can only be $2$ or $5$ . This implies that if $d\mid2$ or $d\mid5$ , the fraction is reducible. There are $249$ cases where $d\mid2$ $99$ where $d\mid5$ , and $49$ where $d\mid(2\cdot5=10)$ , so by PIE, the number of fails is $299$ , so our answer is $\boxed{200}$ | null | 200 |
92fc8a80dff5911d9be553ea46895107 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | We know that the numerator of the fraction cannot be even, because the denominator would also be even. We also know that the numerator cannot be a multiple of $5$ because the denominator would also be a multiple of $5$ . Proceed by listing out all the other possible fractions and we realize that the numerator and denominator are always relatively prime. We have $499$ fractions to start with, and $250$ with odd numerators. Subtract $50$ to account for the multiples of $5$ , and we get $\boxed{200}$ | null | 200 |
92fc8a80dff5911d9be553ea46895107 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | We know that the set of these rational numbers is from $\dfrac{1}{999}$ to $\dfrac{499}{501}$ where each each element $\dfrac{n}{m}$ has $n+m =1000$ and $\dfrac{n}{m}$ is irreducible.
We note that $\dfrac{n}{m} =\dfrac{1000-m}{m}=\dfrac{1000}{m}-1$ .
Hence, $\dfrac{n}{m}$ is irreducible if $\dfrac{1000}{m}$ is irreducible, and $\dfrac{1000}{m}$ is irreducible if $m$ is not divisible by $2$ or $5$ . Thus, the answer to the question is the number of integers between $501$ and $999$ inclusive that are not divisible by $2$ or $5$
We note there are $499$ numbers between $501$ and $999$ , and
Using the Principle of Inclusion and Exclusion, we get that there are $499-249-99+49=200$ numbers between $501$ and $999$ are not divisible by either $2$ or $5$ , so our answer is $\boxed{200}$ | null | 200 |
92fc8a80dff5911d9be553ea46895107 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | We notice that there are a total of $400$ fractions that are in simplest form where the numerator and denominator add up to $1000$ . Because the numerator and denominator have to be relatively prime, there are $\varphi(1000)=400$ fractions. Half of these are greater than $1$ , so the answer is $400\div2=\boxed{200}$ | null | 200 |
92fc8a80dff5911d9be553ea46895107 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | Our fraction can be written in the form $\frac{1000 - a}{a} = \frac{1000}{a} - 1.$ Thus the fraction is reducible when $a$ divides $1000.$ We also want $500 < a < 1000.$ By PIE , the total values of $a$ that make the fraction reducible is, \[249 + 99 - 49 = 299.\] By complementary counting , the answer we want is $499 - 299 = \boxed{200}.$ | null | 200 |
92fc8a80dff5911d9be553ea46895107 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_3 | Find the number of rational numbers $r$ $0<r<1$ , such that when $r$ is written as a fraction in lowest terms, the numerator and the denominator have a sum of 1000. | Suppose our fraction is $\frac{a}{b}$ . The given condition means $a+b=1000$ . Now, if $a$ and $b$ share a common factor greater than $1$ , then the expression $a+b$ must also contain that common factor. This means our fraction cannot have a factor of $5$ or be even.
There are $250$ fractions that aren’t even. From this, $50$ are divisible by $5$ , which means the answer is $250-50=\boxed{200}$ | null | 200 |
594d7d3ddb1c28aab1d89e97b2887dc8 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_4 | Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | For the purposes of this problem, we will use miles and minutes as our units; thus, the bikers travel at speeds of $\dfrac{1}{3}$ mi/min.
Let $d$ be the length of the trains, $r_1$ be the speed of train 1 (the faster train), and $r_2$ be the speed of train 2.
Consider the problem from the bikers' moving frame of reference. In order to pass Jon, the first train has to cover a distance equal to its own length, at a rate of $r_1 - \dfrac{1}{3}$ . Similarly, the second train has to cover a distance equal to its own length, at a rate of $r_2 + \dfrac{1}{3}$ . Since the times are equal and $d = rt$ , we have that $\dfrac{d}{r_1 - \dfrac{1}{3}} = \dfrac{d}{r_2 + \dfrac{1}{3}}$ . Solving for $r_1$ in terms of $r_2$ , we get that $r_1 = r_2 + \dfrac{2}{3}$
Now, let's examine the times it takes the trains to pass Steve. This time, we augment train 1's speed by $\dfrac{1}{3}$ , and decrease train 2's speed by $\dfrac{1}{3}$ . Thus, we have that $\dfrac{d}{r_2 - \dfrac{1}{3}} = 10\dfrac{d}{r_1 + \dfrac{1}{3}}$
Multiplying this out and simplifying, we get that $r_1 = 10r_2 - \dfrac{11}{3}$ . Since we now have 2 expressions for $r_1$ in terms of $r_2$ , we can set them equal to each other:
$r_2 + \dfrac{2}{3} = 10r_2 - \dfrac{11}{3}$ . Solving for $r_2$ , we get that $r_2 = \dfrac{13}{27}$ . Since we know that it took train 2 1 minute to pass Jon, we know that $1 = \dfrac{d}{r_2 + \dfrac{1}{3}}$ . Plugging in $\dfrac{13}{27}$ for $r_2$ and solving for $d$ , we get that $d = \dfrac{22}{27}$ , and our answer is $27 + 22 = \boxed{049}$ | null | 049 |
594d7d3ddb1c28aab1d89e97b2887dc8 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_4 | Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Using a similar approach to Solution 1, let the speed of the east bound train be $a$ and the speed of the west bound train be $b$
So $a-20=b+20$ and $a+20=10(b-20)$
From the first equation, $a=b+40$ . Substituting into the second equation, \[b+60=10b-200\] \[260=9b\] \[b=\frac{260}{9}\text{ mph}\] This means that \[a=\frac{260}{9}+40=\frac{620}{9}\text{ mph}\] Checking, we get that the common difference in Jon's speed and trains' speeds is $\frac{440}{9}$ and the differences for Steve is $\frac{800}{9}$ and $\frac{80}{9}$
This question assumes the trains' lengths in MILES: \[\frac{440}{9}\cdot \frac{1}{60}=\frac{440}{540}=\frac{22}{27}\text{ miles}\] Adding up, we get $22+27=\boxed{049}$ | null | 049 |
594d7d3ddb1c28aab1d89e97b2887dc8 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_4 | Jon and Steve ride their bicycles along a path that parallels two side-by-side train tracks running the east/west direction. Jon rides east at $20$ miles per hour, and Steve rides west at $20$ miles per hour. Two trains of equal length, traveling in opposite directions at constant but different speeds each pass the two riders. Each train takes exactly $1$ minute to go past Jon. The westbound train takes $10$ times as long as the eastbound train to go past Steve. The length of each train is $\tfrac{m}{n}$ miles, where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let the length of the trains be $L$ , let the rate of the westward train be $W_{R}$ , ;et the rate of the eastward train be $E_{R}$ , and let the time it takes for the eastward train to pass Steve be $E_{T}$
We have that
$L=(\frac{1}{60})(W_{R}+20)$
$L=(\frac{1}{60})(E_{R}-20)$
Adding both of the equations together, we get that
$2L=\frac{W_{R}}{60}+\frac{E_{R}}{60}\implies 120L=W_{R}+E_{R}$
Now, from the second part of the problem, we acquire that
$L=(E_{T})(E_{R}+20)$
$L=(10E_{T})(W_{R}-20)$
Dividing the second equation by the first, we get that... $1=\frac{10(W_{R}-20)}{E_{R}+20}\implies E_{R}+20=10W_{R}-200\implies E_{R}+220=10W_{R}\implies E_{R}=10W_{R}-220$
Now, substituting into the $120L=W_{R}+E_{R}$
$120L=W_{R}+(10W_{R}-220)\implies 120L= 11W_{R}-220\implies W_{R}=\frac{120L+220}{11}$
Finally, plugging this back into our very first equation..
$L=(\frac{1}{60})((\frac{120L+220}{11})+20)\implies 660L=120L+440\implies 540L=440\implies L=\frac{22}{27}$
Hence, the answer is $22+27=\boxed{049}$ | null | 049 |
3d7f0a7558d8c6278763faa95740146c | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_5 | Let the set $S = \{P_1, P_2, \dots, P_{12}\}$ consist of the twelve vertices of a regular $12$ -gon. A subset $Q$ of $S$ is called "communal" if there is a circle such that all points of $Q$ are inside the circle, and all points of $S$ not in $Q$ are outside of the circle. How many communal subsets are there? (Note that the empty set is a communal subset.) | By looking at the problem and drawing a few pictures, it quickly becomes obvious that one cannot draw a circle that covers $2$ disjoint areas of the $12$ -gon without including all the vertices in between those areas. In other words, in order for a subset to be communal, all the vertices in the subset must be adjacent to one another. We now count the number of ways to select a row of adjacent vertices. We notice that for any subset size between $1$ and $11$ , there are $12$ possible subsets like this (this is true because we can pick any of the $12$ vertices as a "starting" vertex, include some number of vertices counterclockwise from that vertex, and generate all possible configurations). However, we also have to include the set of all $12$ vertices, as well as the empty set. Thus, the total number is $12\cdot11 + 2 = \boxed{134}$ | null | 134 |
7f04b136348b9ecdbfb7b10822dc3f3d | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6 | The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ | Begin by setting $x$ to 0, then set both equations to $h^2=\frac{2013-j}{3}$ and $h^2=\frac{2014-k}{2}$ , respectively. Notice that because the two parabolas have to have positive x-intercepts, $h\ge32$
We see that $h^2=\frac{2014-k}{2}$ , so we now need to find a positive integer $h$ which has positive integer x-intercepts for both equations.
Notice that if $k=2014-2h^2$ is -2 times a square number, then you have found a value of $h$ for which the second equation has positive x-intercepts. We guess and check $h=36$ to obtain $k=-578=-2(17^2)$
Following this, we check to make sure the first equation also has positive x-intercepts (which it does), so we can conclude the answer is $\boxed{036}$ | null | 036 |
7f04b136348b9ecdbfb7b10822dc3f3d | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6 | The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ | Let $x=0$ and $y=2013$ for the first equation, resulting in $j=2013-3h^2$ . Substituting back in to the original equation, we get $y=3(x-h)^2+2013-3h^2$
Now we set $y$ equal to zero, since there are two distinct positive integer roots. Rearranging, we get $2013=3h^2-3(x-h)^2$ , which simplifies to $671=h^2-(x-h)^2$ . Applying difference of squares, we get $671=(2h-x)(x)$
Now, we know that $x$ and $h$ are both integers, so we can use the fact that $671=61\times11$ , and set $2h-x=11$ and $x=61$ (note that letting $x=11$ gets the same result). Therefore, $h=\boxed{036}$ | null | 036 |
7f04b136348b9ecdbfb7b10822dc3f3d | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6 | The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ | Similar to the first two solutions, we deduce that $\text{(-)}j$ and $\text{(-)}k$ are of the form $3a^2$ and $2b^2$ , respectively, because the roots are integers and so is the $y$ -intercept of both equations. So the $x$ -intercepts should be integers also.
The first parabola gives \[3h^2+j=3\left(h^2-a^2\right)=2013\] \[h^2-a^2=671\] And the second parabola gives \[2h^2+k=2\left(h^2-b^2\right)=2014\] \[h^2-b^2=1007\]
We know that $671=11\cdot 61$ and that $1007=19\cdot 53$ . It is just a fitting coincidence that the average of $11$ and $61$ is the same as the average of $19$ and $53$ . That is $\boxed{036}$ | null | 036 |
7f04b136348b9ecdbfb7b10822dc3f3d | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6 | The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ | First, we expand both equations to get $y=3x^2-6hx+3h^2+j$ and $y=2x^2-4hx+2h^2+k$ . The $y$ -intercept for the first equation can be expressed as $3h^2+j$ . From this, the x-intercepts for the first equation can be written as
\[x=h \pm \sqrt{(-6h)^2-4*3(3h^2+j)}=h \pm \sqrt{36h^2-12(2013)}=h \pm \sqrt{36h^2-24156}\]
Since the $x$ -intercepts must be integers, $\sqrt{36h^2-24156}$ must also be an integer. From solution 1, we know $h$ must be greater than or equal to 32. We can substitute increasing integer values for $h$ starting from 32; we find that $h=36$
We can test this result using the second equation, whose $x$ -intercepts are \[x=h \pm \sqrt{(-4h)^2-4*2(2h^2+k)}=h \pm \sqrt{16h^2-8(2014)}=h \pm \sqrt{16h^2-16112}\] Substituting 36 in for $h$ , we get $h=36 \pm 68$ , which satisfies the requirement that all x-intercepts must be (positive) integers.
Thus, $h=\boxed{036}$ | null | 036 |
7f04b136348b9ecdbfb7b10822dc3f3d | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6 | The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ | We have the equation $y=3(x-h)^2 + j.$
We know: $(x,y):(0,2013)$ , so $h^2=2013/3 - j/3$ after plugging in the values and isolating $h^2$ . Therefore, $h^2=671-j/3$
Lets call the x-intercepts $x_1$ $x_2$ . Since both $x_1$ and $x_2$ are positive there is a relationship between $x_1$ $x_2$ and $h$ . Namely, $x_1+x_2=2h$ . The is because: $x_1-h=-(x_2-h)$
Similarly, we know: $(x,y):(x_1,0)$ , so $j=-3(x_1-h)^2$ . Combining the two equations gives us \[h^2=671+(x_1-h)^2\] \[h^2=671+x_1^2-2x_1h+h^2\] \[h=(671+x_1^2)/2x_1.\]
Now since we have this relationship, $2h=x_1+x_2$ , we can just multiply the last equation by 2(so that we get $2h$ on the left side) which gives us \[2h=671/x_1+x_1^2/x^1\] \[2h=671/x_1+x_1\] \[x_1+x_2=671/x_1+x_1\] \[x_2=671/x_1\] \[x_1x_2=671.\] Prime factorization of 671 gives 11 and 61. So now we know $x_1=11$ and $x_2=61$ . Lastly, we plug in the numbers,11 and 61, into $x_1+x_2=2h$ , so $\boxed{36}$ | null | 36 |
7f04b136348b9ecdbfb7b10822dc3f3d | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_6 | The graphs $y=3(x-h)^2+j$ and $y=2(x-h)^2+k$ have y-intercepts of $2013$ and $2014$ , respectively, and each graph has two positive integer x-intercepts. Find $h$ | First, we start of exactly like solutions above and we find out that $j=2013-3h^2$ and $k=2014-2h^2$ We then plug j and k into $3(x-h)^2+j$ and $y=2(x-h)^2+k$ respectively. After that, we get two equations, $y=3x^2-6xh+2013$ and $y=2x^2-4xh+2014$ . We can apply Vieta's. Let the roots of the first equation be $a, b$ and the roots of the second equation be $c, d$ . Thus, we have that $a\cdot b=1007$ $a+b=2h$ and $c\cdot d=671$ $c+d=2h$ . Simple evaluations finds that $\boxed{36}$ | null | 36 |
09595c89057e8e68df76056419beb46a | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_7 | Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane) | Let $w = \operatorname{cis}{(\alpha)}$ and $z = 10\operatorname{cis}{(\beta)}$ . Then, $\dfrac{w - z}{z} = \dfrac{\operatorname{cis}{(\alpha)} - 10\operatorname{cis}{(\beta)}}{10\operatorname{cis}{\beta}}$
Multiplying both the numerator and denominator of this fraction by $\operatorname{cis}{(-\beta)}$ gives us:
$\dfrac{w - z}{z} = \dfrac{1}{10}\operatorname{cis}{(\alpha - \beta)} - 1 = \dfrac{1}{10}\cos{(\alpha - \beta)} + \dfrac{1}{10}i\sin{(\alpha - \beta)} - 1$
We know that $\tan{\theta}$ is equal to the imaginary part of the above expression divided by the real part. Let $x = \alpha - \beta$ . Then, we have that:
$\tan{\theta} = \dfrac{\sin{x}}{\cos{x} - 10}.$
We need to find a maximum of this expression, so we take the derivative:
Note (not from author): To take the derivative, we need to use the Quotient Rule . In this case, \[\frac{d}{dx}\left(\frac{\sin x}{\cos x-10}\right)=\frac{\cos x(\cos x-10)-(-\sin x)\sin x}{(\cos x-10)^2}=\dfrac{1 - 10\cos{x}}{(\cos{x} - 10)^2}\]
Thus, we see that the maximum occurs when $\cos{x} = \dfrac{1}{10}$ . Therefore, $\sin{x} = \pm\dfrac{\sqrt{99}}{10}$ , and $\tan{\theta} = \pm\dfrac{\sqrt{99}}{99}$ . Thus, the maximum value of $\tan^2{\theta}$ is $\dfrac{99}{99^2}$ , or $\dfrac{1}{99}$ , and our answer is $1 + 99 = \boxed{100}$ | null | 100 |
09595c89057e8e68df76056419beb46a | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_7 | Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane) | Without the loss of generality one can let $z$ lie on the positive x axis and since $arg(\theta)$ is a measure of the angle if $z=10$ then $arg(\dfrac{w-z}{z})=arg(w-z)$ and we can see that the question is equivalent to having a triangle $OAB$ with sides $OA =10$ $AB=1$ and $OB=t$ and trying to maximize the angle $BOA$ [asy] pair O = (0,0); pair A = (100,0); pair B = (80,30); pair D = (sqrt(850),sqrt(850)); draw(A--B--O--cycle); dotfactor = 3; dot("$A$",A,dir(45)); dot("$B$",B,dir(45)); dot("$O$",O,dir(135)); dot("$ \theta$",O,(7,1.2)); label("$1$", ( A--B )); label("$10$",(O--A)); label("$t$",(O--B)); [/asy]
using the Law of Cosines we get: $1^2=10^2+t^2-t*10*2\cos\theta$ rearranging: \[20t\cos\theta=t^2+99\] solving for $\cos\theta$ we get:
\[\frac{99}{20t}+\frac{t}{20}=\cos\theta\] if we want to maximize $\theta$ we need to minimize $\cos\theta$ , using AM-GM inequality we get that the minimum value for $\cos\theta= 2\left(\sqrt{\dfrac{99}{20t}\dfrac{t}{20}}\right)=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}$ hence using the identity $\tan^2\theta=\sec^2\theta-1$ we get $\tan^2\theta=\frac{1}{99}$ and our answer is $1 + 99 = \boxed{100}$ | null | 100 |
09595c89057e8e68df76056419beb46a | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_7 | Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$ . Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$ . The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ . (Note that $\arg(w)$ , for $w \neq 0$ , denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane) | Note that $\frac{w-z}{z}=\frac{w}{z}-1$ , and that $\left|\frac{w}{z}\right|=\frac{1}{10}$ . Thus $\frac{w}{z}-1$ is a complex number on the circle with radius $\frac{1}{10}$ and centered at $-1$ on the complex plane. Let $\omega$ denote this circle.
Let $A$ and $C$ be the points that represent $\frac{w}{z}-1$ and $-1$ respectively on the complex plane. Let $O$ be the origin. In order to maximize $\tan^2(\theta)$ , we need to maximize $\angle{AOC}$ . This angle is maximized when $AO$ is tangent to $\omega$ . Using the Pythagorean Theorem, we get
\[AO^2=1^2-\left(\frac{1}{10}\right)^2=\frac{99}{100}\]
Thus
\[\tan^2(\theta)=\frac{AC^2}{AO^2}=\frac{1/100}{99/100}=\frac{1}{99}\]
And the answer is $1+99=\boxed{100}$ | null | 100 |
6e6a926f5bdf0cf28957b0ce7c89462c | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8 | The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$ | We have that $N^2 - N = N(N - 1)\equiv 0\mod{10000}$
Thus, $N(N-1)$ must be divisible by both $5^4$ and $2^4$ . Note, however, that if either $N$ or $N-1$ has both a $5$ and a $2$ in its factorization, the other must end in either $1$ or $9$ , which is impossible for a number that is divisible by either $2$ or $5$ . Thus, one of them is divisible by $2^4 = 16$ , and the other is divisible by $5^4 = 625$ . Noting that $625 \equiv 1\mod{16}$ , we see that $625$ would work for $N$ , except the thousands digit is $0$ . The other possibility is that $N$ is a multiple of $16$ and $N-1$ is a multiple of $625$ . In order for this to happen, \[N-1 \equiv -1 \pmod {16}.\] Since $625 \equiv 1 \pmod{16}$ , we know that $15 \cdot 625 = 9375 \equiv 15 \equiv -1 \mod{16}$ . Thus, $N-1 = 9375$ , so $N = 9376$ , and our answer is $\boxed{937}$ | null | 937 |
6e6a926f5bdf0cf28957b0ce7c89462c | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8 | The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$ | let $N= 10000t+1000a+100b+10c+d$ for positive integer values $t,a,b,c,d$ .
When we square $N$ we get that \begin{align*} N^2 &=(10000t+1000a+100b+10c+d)^2\\ &=10^8t^2+10^6a^2+10^4b^2+10^2c^2+d^2+2(10^7ta+10^6tb+10^5tc+10^4td+10^5ab+10^4ac+10^3bc+10^ad+10^2bd+10cd) \end{align*}
However, we don't have to deal with this whole expression but only with its last 4 digits so it is suffices to consider only: \[2000ad+2000bc+100c^2+200bd+20cd+d^2.\] Now we need to compare each decimal digit with $1000a+100b+10c+d$ and see whether the digits are congruent in base 10.
we first consider the ones digits:
$d^2\equiv d \pmod{10}.$
This can happen for only 3 values : 1, 5 and 6.
We can try to solve each case:
Considering the tenths place,
we have that:
$20cd=20c\equiv 10c \pmod {100}$ so $c= 0$
Considering the hundreds place we have that
$200bd+100c^2= 200b \equiv 100b \pmod{1000}$ so again $b=0$
now considering the thousands place we have that
$2000ad+2000bc = 2000a \equiv 1000a \pmod {10000}$ so we get $a=0$ but $a$ cannot be equal to $0$ so we consider $d=5.$
considering the tenths place
we have that:
$20cd+20=100c+20\equiv 20 \equiv 10c \mod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$ )
so $c=2$
considering the hundreds place we have that
$200bd+100c^2+100c= 1000b+600 \equiv600\equiv 100b \pmod{1000}$ ( the extra $100c$ is carried from the tenths place)
so $b=6$
now considering the thousands place we have that
$2000ad+2000bc +1000b= 10000a+24000+ 6000\equiv0\equiv 1000a \pmod {10000}$ ( the extra $1000b$ is carried from the hundreds place)
so a is equal 0 again
considering the tenths place
we have that:
$20cd+30=120c+30\equiv 30+20c \equiv 10c \pmod {100}$ ( the extra $20$ is carried from $d^2$ which is equal to $25$ )
if $c=7$ then we have
$30+20 \cdot 7 \equiv 70\equiv7 \cdot 10 \pmod{100}$
so $c=7$
considering the hundreds place we have that
$200bd+100c^2+100c+100= 1200b+4900+800 \equiv200b+700\equiv 100b \pmod{1000}$ ( the extra $100c+100$ is carried from the tenths place)
if $b=3$ then we have
$700+200 \cdot 3 \equiv 300\equiv3 \cdot 100 \pmod {1000}$
so $b=3$
now considering the thousands place we have that
$2000ad+2000bc +1000b+5000+1000= 12000a+42000+ 3000+6000\equiv0\equiv 2000a+1000\equiv 1000a \pmod {10000}$ ( the extra $1000b+6000$ is carried from the hundreds place)
if $a=9$ then we have
$2000 \cdot 9+1000 \equiv 9000\equiv9 \cdot 1000 \pmod {1000}$
so $a=9$
so we have that the last 4 digits of $N$ are $9376$ and $abc$ is equal to $\boxed{937}$ | null | 937 |
6e6a926f5bdf0cf28957b0ce7c89462c | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8 | The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$ | By the Chinese Remainder Theorem, the equation $N(N-1)\equiv 0\pmod{10000}$ is equivalent to the two equations: \begin{align*} N(N-1)&\equiv 0\pmod{16},\\ N(N-1)&\equiv 0\pmod{625}. \end{align*} Since $N$ and $N-1$ are coprime, the only solutions are when $(N\mod{16},N\mod{625})\in\{(0,0),(0,1),(1,0),(1,1)\}$
Let \[\varphi:\mathbb Z/10000\mathbb Z\to\mathbb Z/16\mathbb Z\times\mathbb Z/625\mathbb Z,\] \[x\mapsto (x\mod{16},x\mod{625}).\] The statement of the Chinese Remainder theorem is that $\varphi$ is an isomorphism between the two rings. In this language, the solutions are $\varphi^{-1}(0,0)$ $\varphi^{-1}(0,1)$ $\varphi^{-1}(1,0)$ , and $\varphi^{-1}(1,1)$ . Now we easily see that \[\varphi^{-1}(0,0)=0\] and \[\varphi^{-1}(1,1)=1.\] Noting that $625\equiv 1\pmod{16}$ , it follows that \[\varphi^{-1}(1,0)=625.\] To compute $\varphi^{-1}(0,1)$ , note that \[(0,1)=15(1,0)+(1,1)\] in \[\mathbb Z/16\mathbb Z\times\mathbb Z/625\mathbb Z,\] so since $\varphi^{-1}$ is linear in its arguments (by virtue of being an isomorphism), \[\varphi^{-1}(0,1)=15\varphi^{-1}(1,0)+\varphi^{-1}(1,1)=15\times 625+1=9376.\]
The four candidate digit strings $abcd$ are then $0000,0001,0625,9376$ . Of those, only $9376$ has nonzero first digit, and therefore the answer is $\boxed{937}$ | null | 937 |
6e6a926f5bdf0cf28957b0ce7c89462c | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_8 | The positive integers $N$ and $N^2$ both end in the same sequence of four digits $abcd$ when written in base $10$ , where digit $a$ is not zero. Find the three-digit number $abc$ | WLOG, we can assume that $N$ is a 4-digit integer $\overline{abcd}$ . Note that the only $d$ that will satisfy $N$ will also satisfy $d^2\equiv d\pmod{10}$ , as the units digit of $\overline{abcd}^2$ is affected only by $d$ , regardless of $a$ $b$ , or $c$
By checking the numbers 0-9, we see that the only possible values of $d$ are $d=0, 1, 5, 6$
Now, we seek to find $c$ . Note that the only $\overline{cd}$ that will satisfy $N$ will also satisfy $\overline{cd}^2 \equiv \overline{cd}\pmod{100}$ , by the same reasoning as above - the last two digits of $\overline{abcd}^2$ are only affected by $c$ and $d$ . As we already have narrowed choices for $d$ , we start reasoning out.
First, we note that if $d=0$ , then $c=0$ , as a number ending in 0, and therefore divisible by 10, is squared, the result is divisible by 100, meaning it ends in two 0's. However, if $N$ ends in $00$ , then recursively, $a$ and $b$ must be $0$ , as a number divisible by 100 squared ends in four zeros. As $a$ cannot be 0, we throw out this possibility, as the only solution in this case is $0$
Now, let's assume that $d=1$ $\overline{cd}$ is equal to $10c + d = 10c + 1$ . Squaring this gives $100c^2 + 20c + 1$ , and when modulo 100 is taken, it must equal $10c + 1$ . As $c$ is an integer, $100c^2$ must be divisible by 100, so $100c^2+20c+1 \equiv 20c + 1\pmod{100}$ , which must be equivalent to $10c + 1$ . Note that this is really $\overline{(2c)1}$ and $\overline{c1}$ , and comparing the 10's digits. So really, we're just looking for when the units digit of $2c$ and $c$ are equal, and a quick check reveals that this is only true when $c=0$ .However, if we extend this process to find $b$ and $a$ , we'd find that they are also 0. The only solution in this case is $1$ , and since $a=0$ here, this is not our solution. Therefore, there are no valid solutions in this case.
Let's assume that $d=5$ . Note that $(10c + 5)^2 = 100c^2 + 100c + 25$ , and when modulo $100$ is taken, $25$ is the remainder. So all cases here have squares that end in 25, so $\overline{cd}=25$ is our only case here. A quick check reveals that $25^2=625$ , which works for now.
Now, let $d=6$ . Note that $(10c + 6)^2 = 100c^2 + 120c + 36$ . Taking modulo 100, this reduces to $20c+36$ , which must be equivalent to $10c+6$ . Again, this is similar to $\overline{(2c+3)6}$ and $\overline{c6}$ , so we see when the units digits of $2c+3$ and $c$ are equal. To make checking faster, note that $2c$ is necessarily even, so $2c+3$ is necessarily odd, so $c$ must be odd. Checking all the odds reveals that only $c=3$ works, so this case gives $76$ . Checking quickly $76^2 = 5776$ , which works for now.
Now, we find $b$ , given two possibilities for $\overline{cd}$
Start with $\overline{cd} = 25$ $\overline{bcd} = 100b + \overline{cd} = 100b + 25.$ Note that if we square this, we get $10000b^2 + 5000b + 625$ , which should be equivalent to $100b + 25$ modulo 1000. Note that, since $b$ is an integer, $10000b^2 + 5000 + 625$ simplifies modulo 1000 to $625$ . Therefore, the only $\overline{bcd}$ that works here is $625$ $625^2 = 390625$
Now, assume that $\overline{cd}=76$ . We have $100b + 76$ , and when squared, becomes $10000b^2 + 15200b + 5776$ , which, modulo 1000, should be equivalent to $100b+76$ . Reducing $10000b^2 + 15200b + 5776$ modulo $1000$ gives $200b + 776$ . Using the same technique as before, we must equate the hundreds digit of $\overline{(2b+7)76}$ to $\overline{b76}$ , or equate the units digit of $2b+7$ and $b$ . Since $2b+7$ is necessarily odd, any possible $b$ 's must be odd. A quick check reveals that $b=3$ is the only solution, so we get a solution of $376$ $376^2 = 141376$
Finally, we solve for $a$ . Start with $\overline{bcd}=625$ . We have $1000a + 625$ , which, squared, gives \[1000000a^2 + 1250000a + 390625,\] and reducing modulo 10000 gives simply 625. So $\overline{abcd}=625$ . However, that makes $a=0$ . Therefore, no solutions exist in this case.
We turn to our last case, $\overline{bcd}=376$ . We have \[1000a + 376^2 = 1000000a^2 + 752000a + 141376,\] and reducing modulo $10000$ gives $2000a + 1376$ , which must be equivalent to $1000a + 376$ . So we must have $\overline{(2a+1)376}$ being equivalent to $\overline{a376}$ modulo 1000. So, the units digit of $2a+1$ must be equal to $a$ . Since $2a+1$ is odd, $a$ must be odd. Lo and behold, the only possibility for $a$ is $a=3$ . Therefore, $\overline{abcd}=9376$ , so our answer is $\boxed{937}$ | null | 937 |
524ef631d9906f308a2200a7ba0ecbf5 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9 | Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$ | Substituting $n$ for $2014$ , we get \[\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2\] \[= x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0\] Noting that $nx^2 - 1$ factors as a difference of squares to \[(\sqrt{n}x - 1)(\sqrt{n}x+1)\] we can factor the left side as \[(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))\] This means that $\frac{1}{\sqrt{n}}$ is a root, and the other two roots are the roots of $x^2 - 2\sqrt{n}x - 2$ . Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to $2\sqrt{n}$ , so the positive root must be greater than $2\sqrt{n}$ in order to produce this sum when added to a negative value. Since $0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}$ is clearly true, $x_2 = \frac{1}{\sqrt{2014}}$ and $x_1 + x_3 = 2\sqrt{2014}$ . Multiplying these values together, we find that $x_2(x_1+x_3) = \boxed{002}$ | null | 002 |
524ef631d9906f308a2200a7ba0ecbf5 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9 | Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$ | From Vieta's formulae, we know that \[x_1x_2x_3 = \dfrac{-2}{\sqrt{2014}}\] \[x_1 + x_2 + x_3 = \dfrac{4029}{\sqrt{2014}}\] and \[x_1x_2 + x_2x_3 + x_1x_3 = 0\] Thus, we know that \[x_2(x_1 + x_3) = -x_1x_3\]
Now consider the polynomial with roots $x_1x_2, x_2x_3,$ and $x_1x_3$ . Expanding the polynomial \[(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)\] we get the polynomial \[x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2\] Substituting the values obtained from Vieta's formulae, we find that this polynomial is \[x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}\] We know $x_1x_3$ is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to \[1007x^3 - 4029x - 2 = 0\]
Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the $x^3$ term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that $x = -2$ is a solution. Factoring it out, we get that \[1007x^3 - 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)\] Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, \[x_1x_3 = -2\] so \[-x_1x_3 = \boxed{002}\] and we're done. | null | 002 |
524ef631d9906f308a2200a7ba0ecbf5 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9 | Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$ | Observing the equation, we notice that the coefficient for the middle term $-4029$ is equal to \[-2{\sqrt{2014}}^2-1\]
Also notice that the coefficient for the ${x^3}$ term is $\sqrt{2014}$ . Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the $x$ term of the binomial would have a coefficient of $\sqrt{2014}$ . Similarly, the $x$ term of the trinomial would also have a coefficient of $\sqrt{2014}$ . The factored form of the expression would look something like the following: \[({\sqrt{2014}}x-a)(x^2-n{\sqrt{2014}}x-b)\] where ${a, b,c}$ are all positive integers (because the ${x^2}$ term of the original expression is negative, and the constant term is positive), and \[{ab=2}\]
Multiplying this expression out gives \[{{\sqrt{2014}x^3-(2014n+a)x^2+(an{\sqrt{2014}}-b{\sqrt{2014}})x+ab}}\] Equating this with the original expression gives \[{2014n+a}=-4029\] The only positive integer solutions of this expression is $(n, a)=(1, 2015)$ or $(2, 1)$ . If $(n, a)=(1, 2015)$ then setting ${an{\sqrt{2014}}-b{\sqrt{2014}}}=0$ yields ${b=2015}$ and therefore ${ab=2015^2}$ which clearly isn't equal to $2$ as the constant term. Therefore, $(n, a)=(2, 1)$ and the factored form of the expression is: \[({\sqrt{2014}}x-1)(x^2-2{\sqrt{2014}}x-2)\] Therefore, one of the three roots of the original expression is \[{x=\dfrac{1}{\sqrt{2014}}}\] Using the quadratic formula yields the other two roots as \[{x={\sqrt{2014}}+{\sqrt{2016}}}\] and \[{x={\sqrt{2014}}-{\sqrt{2016}}}\] Arranging the roots in ascending order (in the order $x_1<x_2<x_3$ ), \[{\sqrt{2014}}-{\sqrt{2016}}<\dfrac{1}{\sqrt{2014}}<{\sqrt{2014}}+{\sqrt{2016}}\] Therefore, \[x_2(x_1+x_3)=\dfrac{1}{\sqrt{2014}}{2\sqrt{2014}}=\boxed{002}\] | null | 002 |
524ef631d9906f308a2200a7ba0ecbf5 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9 | Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$ | By Vieta's, we are seeking to find $x_2(x_1+x_3)=x_1x_2+x_2x_3=-x_1x_3=\frac{2}{\sqrt{2014}x_2}$ . Substitute $n=-x_1x_3$ and $x_2=\frac{2}{\sqrt{2014}n}$ . Substituting this back into the original equation, we have $\frac{4}{1007n^3}-\frac{8058}{1007n^2}+2=0$ , so $2n^3-\frac{8058}{1007}n+\frac{4}{1007}=2n^3-\frac{8058n-4}{1007}=0$ . Hence, $8058n-4\equiv 2n-4 \equiv 0 \pmod{1007}$ , and $n\equiv 2\pmod{1007}$ . But since $n\le 999$ because it is our desired answer, the only possible value for $n$ is $\boxed{002}$ BEST PROOOFFFF
Stormersyle & mathleticguyyy | null | 002 |
524ef631d9906f308a2200a7ba0ecbf5 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9 | Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$ | Let $x =\frac{y}{\sqrt{2014}}.$ The original equation simplifies to $\frac{y^3}{2014} -\frac{4029y^2}{2014}+2 = 0 \implies y^3 - 4029y^2 + 4028=0.$ Here we clearly see that $y=1$ is a root. Dividing $y-1$ from the sum we find that $(y-1)(y^2-4028y-4028)=0.$ From simple bounding we see that $y=1$ is the middle root. Therefore $x_{2}(x_{1}+x_{3}) =\frac{1}{\sqrt{2014}} \cdot\frac{4028}{\sqrt{2014}} = \boxed{002}.$ | null | 002 |
524ef631d9906f308a2200a7ba0ecbf5 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_9 | Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$ . Find $x_2(x_1+x_3)$ | We will estimate the roots of the polynomial. (This strategy normally doesn't work on AIME #9, but playing around with a function is often good strategy for getting an intuition for the problem. In this problem, estimation happens to be a valid solution path. It isn't a proof, but given the constraint that the answer is an integer, we can be certain that our answer is correct.)
Let $p(x) = \sqrt{2014}x^3-4029x^2+2=0$ . We start by estimating $p(-1)$ $p(0)$ , and $p(1)$ (A natural first step for function analysis.):
$p(-1)\approx -45-4029+2 \approx -4000$
$p(0) = 2$
$p(1) \approx 45-4029+2 \approx -4000$
We conclude by Intermediate Value Theorem (or just common sense), that there is a root on $(-1, 0)$ and another root on $(0, 1)$
We know that $p(1) < 0$ and that $\lim_{x\to\infty} p(x) = \infty$ . We conclude that the third root is on $(1, \infty)$ . Therefore, $x_1 \in (-1, 0)$ $x_2 \in (0, 1)$ , and $x_3 \in (1, \infty)$
We will estimate $x_3$ . For $x > 1$ , the constant term of $p(x)$ is negligible. We simplify and get $p(x_3) \approx \sqrt{2014}x_{3}^3 - 4029x_{3}^2 = 0$ . Solving for $x_3$ (We can divide by $x_{3}^2$ because we know $x_3 \neq 0$ ), we get $x_3 \approx \frac{4029}{\sqrt{2014}} \approx 2\sqrt{2014} \approx 90$ . We can intuitively bound $x_3$ between $88$ and $92$
We will now estimate $x_1$ and $x_2$ $x_1$ and $x_2$ are close to $0$ . As a result, the $\sqrt{2014}x^3$ term is negligible. We simplify and get $p(x) \approx -4029x^2 + 2 = 0$ . Solving for $x$ , we get $x \approx \pm \sqrt{\frac{2}{4029}} \approx \pm \sqrt{\frac{1}{2014.5}} \approx \pm \frac{1}{45}$ . We can intuitively bound $x_1$ between $-\frac{1}{43}$ and $-\frac{1}{47}$ . Similarly, we can intuitively bound $x_2$ between $\frac{1}{47}$ and $\frac{1}{43}$
We calculate that the minimum possible value of $x_2(x_1 + x_3)$ is $\frac{88-\frac{1}{43}}{47}$ and the maximum possible value is $\frac{92-\frac{1}{47}}{43}$ . The only integer that falls is this range is $\boxed{002}$ | null | 002 |
c4fab9337e3fb6f0df0a4da21cb2f695 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_10 | A disk with radius $1$ is externally tangent to a disk with radius $5$ . Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ$ . That is, if the center of the smaller disk has moved to the point $D$ , and the point on the smaller disk that began at $A$ has now moved to point $B$ , then $\overline{AC}$ is parallel to $\overline{BD}$ . Then $\sin^2(\angle BEA)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] size(150); pair a=(5,0),b=(2,3*sqrt(3)),c=(6,0),d=(3,3*sqrt(3)),e=(0,0); draw(circle(e,5)); draw(circle(c,1)); draw(circle(d,1)); dot(a^^b^^c^^d^^e^^(5/2,5*sqrt(3)/2)); label("$A$",a,W,fontsize(9)); label("$B$",b,NW,fontsize(9)); label("$C$",c,E,fontsize(9)); label("$D$",d,E,fontsize(9)); label("$E$",e,SW,fontsize(9)); label("$F$",(5/2,5*sqrt(3)/2),SSW,fontsize(9)); [/asy]
Let $F$ be the new tangency point of the two disks. The smaller disk rolled along minor arc $\overarc{AF}$ on the larger disk.
Let $\alpha = \angle AEF$ , in radians. The smaller disk must then have rolled along an arc of length $5\alpha$ , since the larger disk has a radius of $5$ . Since all of the points on major arc $\overarc{BF}$ on the smaller disk have come into contact with the larger disk at some point during the rolling, and none of the other points on the smaller disk did, \[\overarc{BF}=\overarc{AF}=5\alpha\]
Since $\overline{AC} || \overline{BD}$ \[\angle BDF \cong \angle FEA\] so the angles of minor arc $\overarc{BF}$ and minor arc $\overarc{AF}$ are equal, so minor arc $\overarc{BF}$ has an angle of $\alpha$ . Since the smaller disk has a radius of $1$ , the length of minor arc $\overarc{BF}$ is $\alpha$ . This means that $5\alpha + \alpha$ equals the circumference of the smaller disk, so $6\alpha = 2\pi$ , or $\alpha = \frac{\pi}{3}$
Now, to find $\sin^2{\angle BEA}$ , we construct $\triangle BDE$ . Also, drop a perpendicular from $D$ to $\overline{EA}$ , and call this point $X$ . Since $\alpha = \frac{\pi}{3}$ and $\angle DXE$ is right, \[DE = 6\] \[EX = 3\] and \[DX = 3\sqrt{3}\]
Now drop a perpendicular from $B$ to $\overline{EA}$ , and call this point $Y$ . Since $\overline{BD} || \overline{EA}$ \[XY = BD = 1\] and \[BY = DX = 3\sqrt{3}\] Thus, we know that \[EY = EX - XY = 3 - 1 = 2\] and by using the Pythagorean Theorem on $\triangle BEY$ , we get that \[BE = \sqrt{31}\] Thus, \[\sin{\angle BEA} = \frac{\sqrt{27}}{\sqrt{31}}\] so \[\sin^2{\angle BEA} = \frac{27}{31}\] and our answer is $27 + 31 = \boxed{058}$ | null | 058 |
c4fab9337e3fb6f0df0a4da21cb2f695 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_10 | A disk with radius $1$ is externally tangent to a disk with radius $5$ . Let $A$ be the point where the disks are tangent, $C$ be the center of the smaller disk, and $E$ be the center of the larger disk. While the larger disk remains fixed, the smaller disk is allowed to roll along the outside of the larger disk until the smaller disk has turned through an angle of $360^\circ$ . That is, if the center of the smaller disk has moved to the point $D$ , and the point on the smaller disk that began at $A$ has now moved to point $B$ , then $\overline{AC}$ is parallel to $\overline{BD}$ . Then $\sin^2(\angle BEA)=\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | First, we determine how far the small circle goes. For the small circle to rotate completely around the circumference, it must rotate $5$ times (the circumference of the small circle is $2\pi$ while the larger one has a circumference of $10\pi$ ) plus the extra rotation the circle gets for rotating around the circle, for a total of $6$ times. Therefore, one rotation will bring point $D$ $60^\circ$ from $C$
Now, draw $\triangle DBE$ , and call \[\angle BED = x^{\circ}\] We know that $\overline{ED}$ is 6, and $\overline{BD}$ is 1. Since $EC || BD$ \[\angle BDE = 60^\circ\]
By the Law of Cosines, \[\overline{BE}^2=36+1-2\times 6\times 1\times \cos{60^\circ} = 36+1-6=31\] and since lengths are positive, \[\overline{BE}=\sqrt{31}\]
By the Law of Sines, we know that \[\frac{1}{\sin{x}}=\frac{\sqrt{31}}{\sin{60^\circ}}\] so \[\sin{x} = \frac{\sin{60^\circ}}{\sqrt{31}} = \frac{\sqrt{93}}{62}\] As $x$ is clearly between $0$ and $90^\circ$ $\cos{x}$ is positive. As $\cos{x}=\sqrt{1-\sin^2{x}}$ \[\cos{x} = \frac{11\sqrt{31}}{62}\]
Now we use the angle sum formula to find the sine of $\angle BEA$ \[\sin 60^\circ\cos x + \cos 60^\circ\sin x = \frac{\sqrt{3}}{2}\frac{11\sqrt{31}}{62}+\frac{1}{2}\frac{\sqrt{93}}{62}\] \[= \frac{11\sqrt{93}+\sqrt{93}}{124} = \frac{12\sqrt{93}}{124} = \frac{3\sqrt{93}}{31} = \frac{3\sqrt{31}\sqrt{3}}{31} = \frac{3\sqrt{3}}{\sqrt{31}}\]
Finally, we square this to get \[\frac{9\times 3}{31}=\frac{27}{31}\] so our answer is $27+31=\boxed{058}$ | null | 058 |
622ad46b79a5762a318ee7475815fc76 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11 | A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Perform the coordinate transformation $(x, y)\rightarrow (x+y, x-y)$ . Then we can see that a movement up, right, left, or down in the old coordinates adds the vectors $\langle 1, -1 \rangle$ $\langle 1, 1 \rangle$ $\langle -1, -1 \rangle$ $\langle -1, 1 \rangle$ respectively. Moreover, the transformation takes the equation $|y| = |x|$ to the union of the x and y axis. Exactly half of the moves go up in the new coordinates, and half of them go down. In order to end up on the x axis, we need to go up thrice and down thrice. The number of possible sequences of up and down moves is the number of permutations of $UUUDDD$ , which is just $\binom63 = 20$ . The probability of any of these sequences happening is $\left(\frac12\right)^6$ . Thus, the probability of ending on the x axis is $\frac{20}{2^6}$ . Similarly, the probability of ending on the y axis is the same.
However, we overcount exactly one case: ending at $(0, 0)$ . Since ending on the x axis and ending on the y axis are independent events, the probability of both is simply $\left(\frac{20}{2^6}\right)^2 = \frac{25}{256}$ . Using PIE, the total probability is $\frac{20}{64} + \frac{20}{64} - \frac{25}{256} = \frac{135}{256}$ , giving an answer of $\boxed{391}$ | null | 391 |
622ad46b79a5762a318ee7475815fc76 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11 | A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We have 4 possible moves: U, D, R, and L. The total number of paths that could be taken is $4^6$ , or $4096$ . There are 4 possible cases that land along the line $y = x$ $x,y = \pm 1; x,y = \pm 2; x,y = \pm 3;$ or $x = y = 0$ . We will count the number of ways to end up at $(1,1), (2,2),$ and $(3,3)$ , multiply them by 4 to account for the other quadrants, and add this to the number of ways to end up at $(0,0)$
Thus, the total number of ways to end up at $(1,1)$ is $300$
Thus, the total number of ways to end up at $(0,0)$ is $400$
Adding these cases together, we get that the total number of ways to end up on $y = x$ is $4(20 + 120 + 300) + 400 = 2160$ . Thus, our probability is $\frac{2160}{4096}$ . When this fraction is fully reduced, it is $\frac{135}{256}$ , so our answer is $135 + 256 = \boxed{391}.$ | null | 391 |
622ad46b79a5762a318ee7475815fc76 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11 | A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | We split this into cases by making a chart. In each row, the entries $(\pm1)$ before the dividing line represent the right or left steps (without regard to order), and the entries after the dividing line represent the up or down steps (again, without regard to order). This table only represents the cases where the ending position $(x,y)$ satisfies $x=y$ \[\begin{array}{ccccccccccccl} \multicolumn{5}{c}{R (+)\qquad L (-)}& |&\multicolumn{5}{c}{U (+)\qquad D (-)}\\ +1&& +1&& +1&| & +1&& +1&& +1\\ +1&& +1&& -1& | & +1&& +1&& -1\\ +1&& -1&& -1& | & +1&& -1&& -1\\ -1 && -1&& -1& | & -1&& -1&& -1\\ \\ +1&& +1&& +1&& -1 &|& +1&& +1\\ +1&& +1&& -1 && -1 &|& +1 && -1\\ +1&& -1&& -1 && -1 &|& -1 && -1 &&(\times 2 \text{ for symmetry by swapping }R-L\text{ and }U-D)\\ \\ +1&& +1 &&+1 &&-1&& -1& |& +1\\ +1&& +1 &&-1&& -1&& -1 &|& -1&& (\times 2\text{ symmetry})\\ \\ +1&& +1 &&+1&& -1&& -1 &&-1&| & (\times2 \text{ symmetry})\\ \end{array}\] Note that to account for the cases when $x=-y$ , we can simply multiply the $U-D$ steps by $-1$ , so for example, the first row would become \[+1 \qquad+1\qquad +1 \ \ \ \ |\ \ \ -1\qquad -1\qquad -1.\] Therefore, we must multiply the number of possibilities in each case by $2$ , except for when $x=y=0$
Now, we compute the number of possibilities for each case. In particular, we must compute the number of $RLUD$ words, where $R$ represents $+1$ to the left of $|$ $L$ represents $-1$ to the left of $|$ $U$ represents $+1$ to the right of $|$ , and $D$ represents $-1$ to the right of $|$ . Using multinomials, we compute the following numbers of possibilities for each case. \[{6\choose 3}\cdot 2+ \frac{6!}{2!2!}\cdot 2 + \frac{6!}{2!2!} \cdot 2 + {6\choose 3} \cdot 2 = 2(20 + 180 + 180 + 20) = 800\] \[\frac{6!}{3!2!}\cdot 2 + \frac{6!}{2!2!} + \frac{6!}{3!2!}\cdot 2 = 120 + 180 + 120 = 420\ (\times2\text{ for symmetry})\] \[\frac{6!}{3!2!} \cdot 2 + \frac{6!}{3!2!} \cdot 2 = 120 + 120 = 240\ (\times2\text{ for symmetry})\] \[{6\choose 3} = 20\ (\times 2\text{ for symmetry})\]
Thus, there are $800 + 840 + 480 + 40 = 2160$ possibilities where $|x|=|y|$ . Because there are $4^6$ total possibilities, the probability is $\frac{2160}{4^6} = \frac{135}{256}$ , so the answer is $\boxed{391}.$ | null | 391 |
622ad46b79a5762a318ee7475815fc76 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_11 | A token starts at the point $(0,0)$ of an $xy$ -coordinate grid and then makes a sequence of six moves. Each move is 1 unit in a direction parallel to one of the coordinate axes. Each move is selected randomly from the four possible directions and independently of the other moves. The probability the token ends at a point on the graph of $|y|=|x|$ is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Denote $(x, y)_n$ the probability that starting from point $(x, y)$ , the token reaches a point on the graph of $|y| = |x|$ in exactly $n$ moves. The problem asks us to find $(0, 0)_6$ . We start by breaking this down: \[(0, 0)_6 = \frac14 \cdot ((0, 1)_5 + (0, -1)_5 + (1, 0)_5 + (-1, 0)_5)\] Notice that by symmetry, $(0, 1)_5 = (0, -1)_5 = (1, 0)_5 = (-1, 0)_5$ , so the equation simplifies to \[(0, 0)_6 = (0, 1)_5\] We now expand $(0, 1)_5$ \[(0, 1)_5 = \frac14 \cdot ((0, 0)_4 + (0, 2)_4 + 2(1, 1)_4)\] First, we find $(0, 0)_4$ \[(0, 0)_4 = (0, 1)_3\] \[(0, 1)_3 = \frac14 \cdot ((0, 0)_2 + (0, 2)_2 + 2(1, 1)_2)\] At this point, we can just count the possibilities to find $(0, 0)_2 = \frac34$ $(0, 2)_2 = \frac{7}{16}$ , and $(1, 1)_2 = \frac58$ . Therefore, \[(0, 1)_3 = \frac14 \cdot (\frac34 + \frac{7}{16} + 2 \cdot \frac58)\] \[(0, 1)_3 = \frac{39}{64}\] Next, we find $(0, 2)_4$ \[(0, 2)_4 = \frac14 \cdot ((0, 1)_3 + (0, 3)_3 + 2(1, 2)_3)\] We already calculated $(0, 1)_3$ , so we just need to find $(0, 3)_3$ and $(1, 2)_3$ \[(0, 3)_3 = \frac14 \cdot ((0, 2)_2 + (0, 4)_2 + 2(1, 3)_2)\] \[(0, 3)_3 = \frac14 \cdot (\frac{7}{16} + 0 + 2 \cdot \frac{1}{4})\] \[(0, 3)_3 = \frac{15}{64}\] \[(1, 2)_3 = \frac14 \cdot ((1, 3)_2 + (1, 1)_2 + (0, 2)_2 + (2, 2)_2)\] \[(1, 2)_3 = \frac14 \cdot (\frac14 + \frac58 + \frac{7}{16} + \frac12)\] \[(1, 2)_3 = \frac{29}{64}\] Therefore, \[(0, 2)_4 = \frac14 \cdot (\frac{39}{64} + \frac{15}{64} + 2 \cdot \frac{29}{64})\] \[(0, 2)_4 = \frac{7}{16}\] Finally, we find $(1, 1)_4$ \[(1, 1)_4 = \frac12 \cdot ((0, 1)_3 + (1, 2)_3)\] \[(1, 1)_4 = \frac12 \cdot (\frac{39}{64} + \frac{29}{64})\] \[(1, 1)_4 = \frac{17}{32}\] Putting it all together, \[(0, 0)_6 = (0, 1)_5 =\frac14 \cdot (\frac{39}{64} + \frac{7}{16} + 2 \cdot \frac{17}{32})\] \[(0, 0)_6 = \frac{135}{256}\] Thus, the answer is $135 + 256 = \boxed{391}$ | null | 391 |
72d174cd1c035dc9e37451daa9bd0ff7 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_12 | Let $A=\{1,2,3,4\}$ , and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$ . The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | The natural way to go is casework. And the natural process is to sort $f$ and $g$ based on range size! Via Pigeonhole Principle, we see that the only real possibilities are: $f 1 g 1; f 1 g 2; f 1 g 3; f 2 g 2$ . Note that the $1, 2$ and $1, 3$ cases are symmetrical and we need just a $*2$ . Note also that the total number of cases is $4^4*4^4=4^8$
$f 1 g 1$ : clearly, we choose one number as the range for $f$ , one for $g$ , yielding $12$ possibilities.
$f 1 g 2$ with symmetry (WLOG $f$ has 1 element): start by selecting numbers for the ranges. This yields $4$ for the one number in $f$ , and $3$ options for the two numbers for $g$ . Afterwards, note that the function with 2 numbers in the range can have $4+6+4=14$ arrangements of these two numbers (1 of one, 3 of the other *2 and 2 of each). Therefore, we have $2*12*14$ possibilities, the 2 from symmetry.
$f 2 g 2$ : no symmetry, still easy! Just note that we have $6$ choices of which numbers go to $f$ and $g$ , and within each, $14*14=196$ choices for the orientation of each of the two numbers. That's $6*196$ possibilities.
$f 1 g 3$ : again, symmetrical (WLOG $f$ has one element): $4$ ways to select the single element for $f$ , and then find the number of ways to distribute the $3$ distinct numbers in the range for $g$ . The only arrangement for the frequency of each number is ${1, 1, 2}$ in some order. Therefore, we have $3$ ways to choose which number is the one represented twice, and then note that there are $12$ ways to arrange these! The number of possibilities in this situation is $2 * 4 * 3 * 12$
Total, divided by $4^8$ , gets $\frac{3 * (1 + 2 * 7^2 + 2^2 * 7 + 2^3 * 3)}{4^7}$ , with numerator $\boxed{453}$ | null | 453 |
72d174cd1c035dc9e37451daa9bd0ff7 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_12 | Let $A=\{1,2,3,4\}$ , and $f$ and $g$ be randomly chosen (not necessarily distinct) functions from $A$ to $A$ . The probability that the range of $f$ and the range of $g$ are disjoint is $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m$ | We note there are $4^4 = 256$ possibilities for each of $f$ and $g$ from $A$ to $A$ since the input of the four values of each function has four options each for an output value.
We proceed with casework to determine the number of possible $f$ with range 1, 2, etc.
There are 4 possibilities: all elements output to 1, 2, 3, or 4.
We have ${{4}\choose {2}} = 6$ ways to choose the two output elements for $f$ . At this point we have two possibilities: either $f$ has 3 of 1 element and 1 of the other, or 2 of each element. In the first case, there are 2 ways to pick the element which there are 3 copies of, and ${{4}\choose {1}} = 4$ ways to rearrange the 4 elements, for a total of $6 * 4 * 2 = 48$ ways for this case. For the second case, there are ${{4}\choose {2}} = 6$ ways to rearrange the 4 elements, for a total of $6 * 6 = 36$ ways for this case. Adding these two, we get a total of $36 + 48 = 84$ total possibilities.
We have ${{4}\choose {3}} = 4$ ways to choose the three output elements for $f$ . We know we must have 2 of 1 element and 1 of each of the others, so there are 3 ways to pick this element. Finally, there are ${{4}\choose{1}}*{{3}\choose{1}} = 12$ ways to rearrange these elements (since we can pick the locations of the 2 single elements in this many ways), and our total is $4 * 3 * 12 = 144$ ways.
Since we know the elements present, we have $4!$ ways to arrange them, or 24 ways.
(To check, $4 + 84 + 144 + 24 = 256$ , which is the total number of possibilities).
We now break $f$ down by cases, and count the number of $g$ whose ranges are disjoint from $f$ 's.
We know that there are 3 possibilities for $g$ with 1 element. Since half the possibilities for $g$ with two elements will contain the element in $f$ , there are $84/2 = 42$ possibilities for $g$ with 2 elements. Since $3/4$ the possibilities for $g$ with 3 elements will contain the element in $f$ , there are $144/4 = 36$ possibilities for $g$ with 3 elements. Clearly, no 4-element range for $g$ is possible, so the total number of ways for this case to happen is $4(3 + 42 + 36) = 324$
We know that there are 2 possibilities for $g$ with 1 element. If $g$ has 2 elements in its range, they are uniquely determined, so the total number of sets with a range of 2 elements that work for $g$ is $84/6 = 14$ . No 3-element or 4-element ranges for $g$ are possible. Thus, the total number of ways for this to happen is $84(2 + 14) = 1344$
In this case, there is only 1 possibility for $g$ - all the output values are the element that does not appear in $f$ 's range. Thus, the total number of ways for this to happen is $144$
We find that the probability of $f$ and $g$ having disjoint ranges is equal to:
$\dfrac{324 + 1344 + 144}{256^2}=\dfrac{1812}{2^{16}}= \dfrac{453}{2^{14}}$
Thus, our final answer is $\boxed{453}$ | null | 453 |
76a32848b17aba63c49356a1c18d19c6 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_13 | On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$
[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy] | $269+275+405+411=1360$ , a multiple of $17$ . In addition, $EG=FH=34$ , which is $17\cdot 2$ .
Therefore, we suspect the square of the "hypotenuse" of a right triangle, corresponding to $EG$ and $FH$ must be a multiple of $17$ . All of these triples are primitive:
\[17=1^2+4^2\] \[34=3^2+5^2\] \[51=\emptyset\] \[68=\emptyset\text{ others}\] \[85=2^2+9^2=6^2+7^2\] \[102=\emptyset\] \[119=\emptyset \dots\]
The sides of the square can only equal the longer leg, or else the lines would have to extend outside of the square. Substituting $EG=FH=34$ \[\sqrt{17}\rightarrow 34\implies 8\sqrt{17}\implies A=\textcolor{red}{1088}\] \[\sqrt{34}\rightarrow 34\implies 5\sqrt{34}\implies A=850\] \[\sqrt{85}\rightarrow 34\implies \{18\sqrt{85}/5,14\sqrt{85}/5\}\implies A=\textcolor{red}{1101.6,666.4}\]
Thus, $\boxed{850}$ is the only valid answer. | null | 850 |
76a32848b17aba63c49356a1c18d19c6 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_13 | On square $ABCD$ , points $E,F,G$ , and $H$ lie on sides $\overline{AB},\overline{BC},\overline{CD},$ and $\overline{DA},$ respectively, so that $\overline{EG} \perp \overline{FH}$ and $EG=FH = 34$ . Segments $\overline{EG}$ and $\overline{FH}$ intersect at a point $P$ , and the areas of the quadrilaterals $AEPH, BFPE, CGPF,$ and $DHPG$ are in the ratio $269:275:405:411.$ Find the area of square $ABCD$
[asy] pair A = (0,sqrt(850)); pair B = (0,0); pair C = (sqrt(850),0); pair D = (sqrt(850),sqrt(850)); draw(A--B--C--D--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); dot("$D$",D,dir(45)); pair H = ((2sqrt(850)-sqrt(306))/6,sqrt(850)); pair F = ((2sqrt(850)+sqrt(306)+7)/6,0); dot("$H$",H,dir(90)); dot("$F$",F,dir(270)); draw(H--F); pair E = (0,(sqrt(850)-6)/2); pair G = (sqrt(850),(sqrt(850)+sqrt(100))/2); dot("$E$",E,dir(180)); dot("$G$",G,dir(0)); draw(E--G); pair P = extension(H,F,E,G); dot("$P$",P,dir(60)); label("$w$", intersectionpoint( A--P, E--H )); label("$x$", intersectionpoint( B--P, E--F )); label("$y$", intersectionpoint( C--P, G--F )); label("$z$", intersectionpoint( D--P, G--H ));[/asy] | Continue in the same way as solution 1 to get that $POK$ has area $3a$ , and $OK = \frac{d}{10}$ . You can then find $PK$ has length $\frac 32$
Then, if we drop a perpendicular from $H$ to $BC$ at $L$ , We get $\triangle HLF \sim \triangle OPK$
Thus, $LF = \frac{15\cdot 34}{d}$ , and we know $HL = d$ , and $HF = 34$ . Thus, we can set up an equation in terms of $d$ using the Pythagorean theorem.
\[\frac{15^2 \cdot 34^2}{d^2} + d^2 = 34^2\]
\[d^4 - 34^2 d^2 + 15^2 \cdot 34^2 = 0\]
\[(d^2 - 34 \cdot 25)(d^2 - 34 \cdot 9) = 0\]
$d^2 = 34 \cdot 9$ is extraneous, so $d^2 = 34 \cdot 25$ . Since the area is $d^2$ , we have it is equal to $34 \cdot 25 = \boxed{850}$ | null | 850 |
ca5dcbaba4a87e2854e0f37836fb4cf6 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_14 | Let $m$ be the largest real solution to the equation
$\frac{3}{x-3}+\frac{5}{x-5}+\frac{17}{x-17}+\frac{19}{x-19}=x^2-11x-4$
There are positive integers $a$ $b$ , and $c$ such that $m=a+\sqrt{b+\sqrt{c}}$ . Find $a+b+c$ | The first step is to notice that the 4 on the right hand side can simplify the terms on the left hand side. If we distribute 1 to $\frac{3}{x-3}$ , then the fraction becomes of the form $\frac{x}{x - 3}$ . A similar cancellation happens with the other four terms. If we assume $x = 0$ is not the highest solution (which is true given the answer format) we can cancel the common factor of $x$ from both sides of the equation.
$\frac{1}{x - 3} + \frac{1}{x - 5} + \frac{1}{x - 17} + \frac{1}{x - 19} = x - 11$
Then, if we make the substitution $y = x - 11$ , we can further simplify.
$\frac{1}{y + 8} + \frac{1}{y + 6} + \frac{1}{y - 6} + \frac{1}{y - 8} =y$
If we group and combine the terms of the form $y - n$ and $y + n$ , we get this equation:
$\frac{2y}{y^2 - 64} + \frac{2y}{y^2 - 36} = y$
Then, we can cancel out a $y$ from both sides, knowing that $x = 11$ is not a possible solution given the answer format.
After we do that, we can make the final substitution $z = y^2$
$\frac{2}{z - 64} + \frac{2}{z - 36} = 1$
$2z - 128 + 2z - 72 = (z - 64)(z - 36)$
$4z - 200 = z^2 - 100z + 64(36)$
$z^2 - 104z + 2504 = 0$
Using the quadratic formula, we get that the largest solution for $z$ is $z = 52 + 10\sqrt{2}$ . Then, repeatedly substituting backwards, we find that the largest value of $x$ is $11 + \sqrt{52 + \sqrt{200}}$ . The answer is thus $11 + 52 + 200 = \boxed{263}$ | null | 263 |
0897f869a1c983110ca91919c4002bab | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_15 | In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$ | Since $\angle DBE = 90^\circ$ $DE$ is the diameter of $\omega$ . Then $\angle DFE=\angle DGE=90^\circ$ . But $DF=FE$ , so $\triangle DEF$ is a 45-45-90 triangle. Letting $DG=3x$ , we have that $EG=4x$ $DE=5x$ , and $DF=EF=\frac{5x}{\sqrt{2}}$
Note that $\triangle DGE \sim \triangle ABC$ by SAS similarity, so $\angle BAC = \angle GDE$ and $\angle ACB = \angle DEG$ . Since $DEFG$ is a cyclic quadrilateral, $\angle BAC = \angle GDE=180^\circ-\angle EFG = \angle AFE$ and $\angle ACB = \angle DEG = \angle GFD$ , implying that $\triangle AFE$ and $\triangle CDF$ are isosceles. As a result, $AE=CD=\frac{5x}{\sqrt{2}}$ , so $BE=3-\frac{5x}{\sqrt{2}}$ and $BD =4-\frac{5x}{\sqrt{2}}$
Finally, using the Pythagorean Theorem on $\triangle BDE$ \[\left(3-\frac{5x}{\sqrt{2}}\right)^2 + \left(4-\frac{5x}{\sqrt{2}}\right)^2 = (5x)^2\] Solving for $x$ , we get that $x=\frac{5\sqrt{2}}{14}$ , so $DE= 5x =\frac{25 \sqrt{2}}{14}$ . Thus, the answer is $25+2+14=\boxed{041}$ | null | 041 |
0897f869a1c983110ca91919c4002bab | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_15 | In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$ | [asy] pair A = (0,3); pair B = (0,0); pair C = (4,0); draw(A--B--C--cycle); dotfactor = 3; dot("$A$",A,dir(135)); dot("$B$",B,dir(215)); dot("$C$",C,dir(305)); pair D = (2.21, 0); pair E = (0, 1.21); pair F = (1.71, 1.71); pair G = (2, 1.5); dot("$D$",D,dir(270)); dot("$E$",E,dir(180)); dot("$F$",F,dir(90)); dot("$G$",G,dir(0)); draw(Circle((1.109, 0.609), 1.28)); draw(D--E); draw(E--F); draw(D--F); draw(E--G); draw(D--G); draw(B--F); draw(B--G); [/asy]
First we note that $\triangle DEF$ is an isosceles right triangle with hypotenuse $\overline{DE}$ the same as the diameter of $\omega$ . We also note that $\triangle DGE \sim \triangle ABC$ since $\angle EGD$ is a right angle and the ratios of the sides are $3:4:5$
From congruent arc intersections, we know that $\angle GED \cong \angle GBC$ , and that from similar triangles $\angle GED$ is also congruent to $\angle GCB$ . Thus, $\triangle BGC$ is an isosceles triangle with $BG = GC$ , so $G$ is the midpoint of $\overline{AC}$ and $AG = GC = 5/2$ . Similarly, we can find from angle chasing that $\angle ABF = \angle EDF = \frac{\pi}4$ . Therefore, $\overline{BF}$ is the angle bisector of $\angle B$ . From the angle bisector theorem, we have $\frac{AF}{AB} = \frac{CF}{CB}$ , so $AF = 15/7$ and $CF = 20/7$
Lastly, we apply power of a point from points $A$ and $C$ with respect to $\omega$ and have $AE \times AB=AF \times AG$ and $CD \times CB=CG \times CF$ , so we can compute that $EB = \frac{17}{14}$ and $DB = \frac{31}{14}$ . From the Pythagorean Theorem, we result in $DE = \frac{25 \sqrt{2}}{14}$ , so $a+b+c=25+2+14= \boxed{041}$ | null | 041 |
0897f869a1c983110ca91919c4002bab | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_15 | In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$ | Call $DE=x$ and as a result $DF=EF=\frac{x\sqrt{2}}{2}, EG=\frac{4x}{5}, GD=\frac{3x}{5}$ . Since $EFGD$ is cyclic we just need to get $DG$ and using LoS(for more detail see the $2$ nd paragraph of Solution $2$ ) we get $AG=\frac{5}{2}$ and using a similar argument(use LoS again) and subtracting you get $FG=\frac{5}{14}$ so you can use Ptolemy to get $x=\frac{25\sqrt{2}}{14} \implies \boxed{041}$ .
~First | null | 041 |
0897f869a1c983110ca91919c4002bab | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_15 | In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$ | See inside the $\triangle DEF$ , we can find that $AG>AF$ since if $AG<AF$ , we can see that Ptolemy Theorem inside cyclic quadrilateral $EFGD$ doesn't work. Now let's see when $AG>AF$ , since $\frac{DG}{EG} = \frac{3}{4}$ , we can assume that $EG=4x;GD=3x;ED=5x$ , since we know $EF=FD$ so $\triangle EFD$ is isosceles right triangle. We can denote $DF=EF=\frac{5x\sqrt{2}}{2}$ .Applying Ptolemy Theorem inside the cyclic quadrilateral $EFGD$ we can get the length of $FG$ can be represented as $\frac{x\sqrt{2}}{2}$ . After observing, we can see $\angle AFE=\angle EDG$ , whereas $\angle A=\angle EDG$ so we can see $\triangle AEF$ is isosceles triangle. Since $\triangle ABC$ is a $3-4-5$ triangle so we can directly know that the length of AF can be written in the form of $3x\sqrt{2}$ . Denoting a point $J$ on side $AC$ with that $DJ$ is perpendicular to side $AC$ . Now with the same reason, we can see that $\triangle DJG$ is a isosceles right triangle, so we can get $GJ=\frac{3x\sqrt{2}}{2}$ while the segment $CJ$ is $2x\sqrt{2}$ since its 3-4-5 again. Now adding all those segments together we can find that $AC=5=7x\sqrt{2}$ and $x=\frac{5\sqrt{2}}{14}$ and the desired $ED=5x=\frac{25\sqrt{2}}{14}$ which our answer is $\boxed{041}$ ~bluesoul | null | 041 |
0897f869a1c983110ca91919c4002bab | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_15 | In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$ | The main element of the solution is the proof that $BF$ is bisector of $\angle B.$
Let $O$ be the midpoint of $DE.$ $\angle EBF = 90^\circ \implies$
$O$ is the center of the circle $BDGFE.$ $\angle EOF = 90^\circ \implies \overset{\Large\frown} {EF} = 90^\circ \implies \angle EBF = 45^\circ \implies$ BF is bisector of $\angle ABC\implies BF = \frac {2AB \cdot BC}{AB+BC} \cos 45^\circ =\frac {12 \cdot \sqrt{2}}{7}.$ \[\angle EGD = 90^\circ, \frac {EG}{GD}=\frac{4}{3} \implies\] \[\angle GED = \angle GCD =\gamma \implies \overset{\Large\frown} {DG} = 2\gamma.\] \[2\angle ACB = \overset{\Large\frown} {BEF} - \overset{\Large\frown} {DG} \implies \overset{\Large\frown} {BEF} = 4 \gamma \implies\] \[\angle BOF = 4 \gamma \implies \angle OBF = \angle OFB = 90^\circ – 2 \gamma.\] Let $BO = EO = DO = r \implies BF = 2 r \cos(90^\circ – 2\gamma) =$ \[=2 r \sin 2\gamma = 4r \sin \gamma \cdot \cos \gamma = 4 r\cdot \frac {3}{5} \cdot \frac {4}{5} = \frac {48}{25} = \frac {12 \cdot \sqrt{2}}{7}\implies\] \[r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = 2r = \frac {25 \cdot \sqrt{2}}{14}\implies \boxed{041}.\] vladimir.shelomovskii@gmail.com, vvsss | null | 041 |
0897f869a1c983110ca91919c4002bab | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_15 | In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$ | The main element of the solution is the proof that $G$ is midpoint of $AC.$
As in Solution 5 we get $\angle GED = \angle DBG =\gamma \implies$
$\triangle BCG$ is isosceles triangle with $BG=CG.$
Similarly $BG = AG \implies AG = CG = BG = \frac {AC}{2} =\frac {5}{2}.$
\[\overset{\Large\frown} {FG} = 90^\circ – \overset{\Large\frown} {GD} = 90^\circ – 2\gamma \implies\] \[\overset{\Large\frown} {BFG} = 4\gamma + 90^\circ – 2\gamma = 90^\circ + 2\gamma \implies\] \[\angle BOG = 90^\circ + 2\gamma \implies \angle BGO = \angle GBO = 45^\circ - \gamma.\] Let $\hspace{10mm} BO = EO = DO = r \implies$ \[BG = 2 r \cos(45^\circ – \gamma) = 2 r (\sin \gamma + \cos \gamma)\frac {\sqrt {2}}{2} =\] \[r \biggl(\frac {3}{5} + \frac {4}{5}\biggr) \sqrt {2} = r \frac {7 \sqrt{2}}{5} = \frac {5}{2}\implies\] \[r = \frac {25 \cdot \sqrt{2}}{28}\implies ED = \frac {25 \cdot \sqrt{2}}{14}\implies \boxed{041}.\] vladimir.shelomovskii@gmail.com, vvsss | null | 041 |
0897f869a1c983110ca91919c4002bab | https://artofproblemsolving.com/wiki/index.php/2014_AIME_I_Problems/Problem_15 | In $\triangle ABC$ $AB = 3$ $BC = 4$ , and $CA = 5$ . Circle $\omega$ intersects $\overline{AB}$ at $E$ and $B$ $\overline{BC}$ at $B$ and $D$ , and $\overline{AC}$ at $F$ and $G$ . Given that $EF=DF$ and $\frac{DG}{EG} = \frac{3}{4}$ , length $DE=\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. Find $a+b+c$ | Let $(BEFGD) = \omega$ .
By Incenter-Excenter(Fact $5$ ), $F$ is the angle bisector of $\angle B$ .
Then by Ratio Lemma we have \[\frac{AG}{CG} = \frac{\sin(ABG)}{\sin(CBG)} \cdot \frac{AB}{BC} = \frac{\sin(GDE)}{\sin(DEG)} \cdot \frac{3}{4} = 1\] Thus, $G$ is the midpoint of $AC$
We can calculate $AF$ and $CF$ to be $\frac{15}{7}$ and $\frac{20}{7}$ respectively.
And then by Power of a Point, we have $\newline$ \[\operatorname{Pow}_{\omega}(A) = AE \cdot AB = AF \cdot AG \implies AE = \frac{25}{14}\] And then similarly, we have $CD = AE = \frac{25}{14}$ $\newline$
Then $EB = \frac{17}{14}$ and $DB = \frac{31}{14}$ and by Pythagorean we have $DE = \frac{25\sqrt{2}}{14}$ , so our answer is $\boxed{041}.$ | null | 041 |
eb1bb46d6d35a09e421396cae6d759d4 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_1 | Abe can paint the room in 15 hours, Bea can paint 50 percent faster than Abe, and Coe can paint twice as fast as Abe. Abe begins to paint the room and works alone for the first hour and a half. Then Bea joins Abe, and they work together until half the room is painted. Then Coe joins Abe and Bea, and they work together until the entire room is painted. Find the number of minutes after Abe begins for the three of them to finish painting the room. | From the given information, we can see that Abe can paint $\frac{1}{15}$ of the room in an hour, Bea can paint $\frac{1}{15}\times\frac{3}{2} = \frac{1}{10}$ of the room in an hour, and Coe can paint the room in $\frac{1}{15}\times 2 = \frac{2}{15}$ of the room in an hour. After $90$ minutes, Abe has painted $\frac{1}{15}\times\frac{3}{2}=\frac{1}{10}$ of the room. Working together, Abe and Bea can paint $\frac{1}{15}+\frac{1}{10}=\frac{1}{6}$ of the room in an hour, so it takes then $\frac{2}{5}\div \frac{1}{6}= \frac{12}{5}$ hours to finish the first half of the room. All three working together can paint $\frac{1}{6}+\frac{2}{15}=\frac{3}{10}$ of the room in an hour, and it takes them $\frac{1}{2}\div \frac{3}{10}=\frac{5}{3}$ hours to finish the room. The total amount of time they take is \[90+\frac{12}{5}\times 60+\frac{5}{3}\times 60 = 90+ 144 + 100 = \boxed{334}\] | null | 334 |
2556b9dde7705fe133056b12d7e7f5ca | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_2 | Arnold is studying the prevalence of three health risk factors, denoted by A, B, and C, within a population of men. For each of the three factors, the probability that a randomly selected man in the population has only this risk factor (and none of the others) is 0.1. For any two of the three factors, the probability that a randomly selected man has exactly these two risk factors (but not the third) is 0.14. The probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac{1}{3}$ . The probability that a man has none of the three risk factors given that he does not have risk factor A is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | We first assume a population of $100$ to facilitate solving. Then we simply organize the statistics given into a Venn diagram.
[asy] pair A,B,C,D,E,F,G; A=(0,55); B=(60,55); C=(60,0); D=(0,0); draw(A--B--C--D--A); E=(30,35); F=(20,20); G=(40,20); draw(circle(E,15)); draw(circle(F,15)); draw(circle(G,15)); draw("$A$",(30,52)); draw("$B$",(7,7)); draw("$C$",(53,7)); draw("100",(5,60)); draw("10",(30,40)); draw("10",(15,15)); draw("10",(45,15)); draw("14",(30,16)); draw("14",(38,29)); draw("14",(22,29)); draw("$x$",(30,25)); draw("$y$",(10,45)); [/asy]
Let $x$ be the number of men with all three risk factors. Since "the probability that a randomly selected man has all three risk factors, given that he has A and B is $\frac{1}{3}$ ," we can tell that $x = \frac{1}{3}(x+14)$ , since there are $x$ people with all three factors and 14 with only A and B. Thus $x=7$
Let $y$ be the number of men with no risk factors. It now follows that \[y= 100 - 3 \cdot 10 - 3 \cdot 14 - 7 = 21.\] The number of men with risk factor A is $10+2 \cdot 14+7 = 45$ (10 with only A, 28 with A and one of the others, and 7 with all three). Thus the number of men without risk factor $A$ is 55, so the desired conditional probability is $21/55$ . So the answer is $21+55=\boxed{076}$ | null | 076 |
d924c8b42654253fadbdf7dd9b3d3223 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_3 | A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a^2$
[asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy] | When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below.
[asy] pair R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); draw(T--R,red); draw(X--Z,red); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy]
By Heron's Formula, the area of each isosceles triangle is $\sqrt{(30)(12)(12)(6)}=\sqrt{180\times 12^2}=72\sqrt{5}$ . So the area of both is $144\sqrt{5}$ . From the rectangle, our original area is $36a$ . The area of the rectangle in the hexagon is $24a$ . So we have \[24a+144\sqrt{5}=36a\implies 12a=144\sqrt{5}\implies a=12\sqrt{5}\implies a^2=\boxed{720}.\] | null | 720 |
d924c8b42654253fadbdf7dd9b3d3223 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_3 | A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a^2$
[asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy] | Alternatively, use basic geometry. First, scale everything down by dividing everything by 6. Let $a/6=p$ . Then, the dimensions of the central rectangle in the hexagon is p x 4, and the original rectangle is 6 x p. By Pythagorean theorem and splitting the end triangles of the hexagon into two right triangles, the altitude of the end triangles is $\sqrt{3^2-2^2}=\sqrt{5}$ given 2 as the base of the constituent right triangles. The two end triangles form a large rectangle of area $\sqrt{5}$ $4$ . Then, the area of the hexagon is $4p+4\sqrt{5}$ , and the area of the rectangle is $6p$ . Equating them, $p=2\sqrt{5}$ . Multiply by scale factor of 6 and square it to get $36(20)= 720 \implies a^2=\boxed{720}$ | null | 720 |
49afec12373c8e4235bc9529d72369a0 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4 | The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ | Notice repeating decimals can be written as the following:
$0.\overline{ab}=\frac{10a+b}{99}$
$0.\overline{abc}=\frac{100a+10b+c}{999}$
where a,b,c are the digits. Now we plug this back into the original fraction:
$\frac{10a+b}{99}+\frac{100a+10b+c}{999}=\frac{33}{37}$
Multiply both sides by $999*99.$ This helps simplify the right side as well because $999=111*9=37*3*9$
$9990a+999b+9900a+990b+99c=33/37*37*3*9*99=33*3*9*99$
Dividing both sides by $9$ and simplifying gives:
$2210a+221b+11c=99^2=9801$
At this point, seeing the $221$ factor common to both a and b is crucial to simplify. This is because taking $mod 221$ to both sides results in:
$2210a+221b+11c \equiv 9801 \mod 221 \iff 11c \equiv 77 \mod 221$
Notice that we arrived to the result $9801 \equiv 77 \mod 221$ by simply dividing $9801$ by $221$ and seeing $9801=44*221+77.$ Okay, now it's pretty clear to divide both sides by $11$ in the modular equation but we have to worry about $221$ being multiple of $11.$ Well, $220$ is a multiple of $11$ so clearly, $221$ couldn't be. Also, $221=13*17.$ Now finally we simplify and get:
$c \equiv 7 \mod 221$
But we know $c$ is between $0$ and $9$ because it is a digit, so $c$ must be $7.$ Now it is straightforward from here to find $a$ and $b$
$2210a+221b+11(7)=9801 \iff 221(10a+b)=9724 \iff 10a+b=44$
and since a and b are both between $0$ and $9$ , we have $a=b=4$ . Finally we have the $3$ digit integer $\boxed{447}$ | null | 447 |
49afec12373c8e4235bc9529d72369a0 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4 | The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ | Note that $\frac{33}{37}=\frac{891}{999} = 0.\overline{891}$ . Also note that the period of $0.abab\overline{ab}+0.abcabc\overline{abc}$ is at most $6$ . Therefore, we only need to worry about the sum $0.ababab+ 0.abcabc$ . Adding the two, we get \[\begin{array}{ccccccc}&a&b&a&b&a&b\\ +&a&b&c&a&b&c\\ \hline &8&9&1&8&9&1\end{array}\] From this, we can see that $a=4$ $b=4$ , and $c=7$ , so our desired answer is $\boxed{447}$ | null | 447 |
49afec12373c8e4235bc9529d72369a0 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4 | The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ | Noting as above that $0.\overline{ab} = \frac{10a + b}{99}$ and $0.\overline{abc} = \frac{100a + 10b + c}{999}$ , let $u = 10a + b$ .
Then \[\frac{u}{99} + \frac{10u + c}{999} = \frac{33}{37}\]
\[\frac{u}{11} + \frac{10u + c}{111} = \frac{9\cdot 33}{37}\]
\[\frac{221u + 11c}{11\cdot 111} = \frac{9\cdot 33}{37}\]
\[221u + 11c = \frac{9\cdot 33\cdot 11\cdot 111}{37}\]
\[221u + 11c = 9\cdot 33^2.\]
Solving for $c$ gives
\[c = 3\cdot 9\cdot 33 - \frac{221u}{11}\]
\[c = 891 - \frac{221u}{11}\]
Because $c$ must be integer, it follows that $u$ must be a multiple of $11$ (because $221$ clearly is not). Inspecting the equation, one finds that only $u = 44$ yields a digit $c, 7$ . Thus $abc = 10u + c = \boxed{447}.$ | null | 447 |
49afec12373c8e4235bc9529d72369a0 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_4 | The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy
\[0.abab\overline{ab}+0.abcabc\overline{abc}=\frac{33}{37},\]
where $a$ $b$ , and $c$ are (not necessarily distinct) digits. Find the three digit number $abc$ | We note as above that $0.\overline{ab} = \frac{10a + b}{99}$ and $0.\overline{abc} = \frac{100a + 10b + c}{999},$ so
\[\frac{10a + b}{99} + \frac{100a + 10b + c}{999} = \frac{33}{37} = \frac{891}{999}.\]
As $\frac{10a + b}{99}$ has a factor of $11$ in the denominator while the other two fractions don't, we need that $11$ to cancel, so $11$ divides $10a + b.$ It follows that $a = b,$ so $\frac{10a + b}{99} = \frac{11a}{99} = \frac{111a}{999},$ so
\[\frac{111a}{999} + \frac{110a+c}{999} = \frac{891}{999}.\]
Then $111a + 110a + c = 891,$ or $221a + c = 891.$ Thus $a = b = 4$ and $c = 7,$ so the three-digit integer $abc$ is $\boxed{447}.$ | null | 447 |
bec47f8f682ab9484242a5373c61c500 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5 | Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ | Because the coefficient of $x^2$ in both $p(x)$ and $q(x)$ is 0, the remaining root of $p(x)$ is $-(r+s)$ , and the remaining root of $q(x)$ is $-(r+s+1)$ . The coefficients of $x$ in $p(x)$ and $q(x)$ are both equal to $a$ , and equating the two coefficients gives \[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2\] from which $s = \tfrac 12 (5r+13)$ . The product of the roots of $p(x)$ differs from that of $q(x)$ by $240$ , so \[(r+4)\cdot \tfrac 12 (5r+7)\cdot \tfrac 12(7r+15)- r\cdot \tfrac 12 (5r+13)\cdot \tfrac 12(7r+13)=240\] from which $r^2+4r-5 =0$ , with roots $r=1$ and $r=-5$
If $r = 1$ , then the roots of $p(x)$ are $r=1$ $s=9$ , and $-(r+s)=-10$ , and $b=-rst=90$
If $r = -5$ , then the roots of $p(x)$ are $r=-5$ $s=-6$ , and $-(r+s)=11$ , and $b=-rst=-330$
Therefore the requested sum is $|- 330| + |90| = \boxed{420}$ | null | 420 |
bec47f8f682ab9484242a5373c61c500 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5 | Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ | Let $r$ $s$ , and $-r-s$ be the roots of $p(x)$ (per Vieta's). Then $r^3 + ar + b = 0$ and similarly for $s$ . Also, \[q(r+4) = (r+4)^3 + a(r+4) + b + 240 = 12r^2 + 48r + 304 + 4a = 0\]
Set up a similar equation for $s$
\[q(s-3) = (s-3)^3 + a(s-3) + b + 240 = -9s^2 + 27s + 213 - 3a = 0.\]
Simplifying and adding the equations gives \begin{align}\tag{*} r^2 - s^2 + 4r + 3s + 49 &= 0 \end{align} Now, let's deal with the $ax$ terms. Plugging the roots $r$ $s$ , and $-r-s$ into $p(x)$ yields a long polynomial, and plugging the roots $r+4$ $s-3$ , and $-1-r-s$ into $q(x)$ yields another long polynomial. Equating the coefficients of $x$ in both polynomials, we get: \[rs + (-r-s)(r+s) = (r+4)(s-3) + (-r-s-1)(r+s+1),\] which eventually simplifies to \[s = \frac{13 + 5r}{2}.\] Substitution into (*) should give $r = -5$ and $r = 1$ , corresponding to $s = -6$ and $s = 9$ , and $|b| = 330, 90$ , for an answer of $\boxed{420}$ | null | 420 |
bec47f8f682ab9484242a5373c61c500 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5 | Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ | The roots of $p(x)$ are $r$ $s$ , and $-r-s$ since they sum to $0$ by Vieta's Formula (co-efficient of $x^2$ term is $0$ ).
Similarly, the roots of $q(x)$ are $r + 4$ $s - 3$ , and $-r-s-1$ , as they too sum to $0$
Then:
$a = rs+r(-r-s)+s(-r-s) = rs-(r+s)^2$ and $-b = rs(-r-s)$ from $p(x)$ and
$a=(r+4)(s-3)+(r+4)(-r-s-1)+(s-3)(-r-s-1) = (r+4)(s-3)-(r+s+1)^2$ and $-(b+240)=(r+4)(s-3)(-r-s-1)$ from $q(x)$
From these equations, we can write that \[rs-(r+s)^2 = (r+4)(s-3)-(r+s+1)^2 = a\] and simplifying gives \[2s-5r-13=0 \Rightarrow s = \frac{5r+13}{2}.\]
We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get \[rs(r+s) = b\] \[(r+4)(s-3)(r+s+1)=b + 240.\] Subtracting the first equation from the second equation gives us $(r+4)(s-3)(r+s+1) - rs(r+s) = 240$
Expanding, simplifying, substituting $s = \frac{5r+13}{2}$ , and simplifying some more yields the simple quadratic $r^2 + 4r - 5 = 0$ , so $r = -5, 1$ . Then $s = -6, 9$
Finally, we substitute back into $b=rs(r+s)$ to get $b = (-5)(-6)(-5-6) = -330$ , or $b = (1)(9)(1 + 9) = 90$
The answer is $|-330|+|90| = \boxed{420}$ | null | 420 |
bec47f8f682ab9484242a5373c61c500 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_5 | Real numbers $r$ and $s$ are roots of $p(x)=x^3+ax+b$ , and $r+4$ and $s-3$ are roots of $q(x)=x^3+ax+b+240$ . Find the sum of all possible values of $|b|$ | By Vieta's, we know that the sum of roots of $p(x)$ is $0$ . Therefore,
the roots of $p$ are $r, s, -r-s$ . By similar reasoning, the roots of $q(x)$ are $r + 4, s - 3, -r - s - 1$ . Thus, $p(x) = (x - r)(x - s)(x + r + s)$ and $q(x) = (x - r - 4)(x - s + 3)(x + r + s + 1)$
Since $p(x)$ and $q(x)$ have the same coefficient for $x$ , we can go ahead
and match those up to get \begin{align*} rs - r(r + s) - s(r + s) &= (r + 4)(s - 3) - (r + 4)(r + s + 1) - (s - 3)(r + s + 1) \\ 0 &= -13 - 5r + 2s \\ s &= \frac{5r + 13}{2} \end{align*}
At this point, we can go ahead and compare the constant term in $p(x)$ and $q(x)$ . Doing so is certainly valid, but we can actually do this another way. Notice that $p(s) = 0$ . Therefore, $q(s) = 240$ . If we plug that into
our expression, we get that \begin{align*} q(s) &= 3(s - r - 4)(r + 2s + 1) \\ 240 &= 3(s - r - 4)(r + 2s + 1) \\ 240 &= 3\left( \frac{3r + 5}{2} \right)(6r + 14) \\ 80 &= (3r + 5)(3r + 7) \\ 0 &= r^2 + 4r - 5 \end{align*} This tells us that $(r, s) = (1, 9)$ or $(-5, -6)$ . Since $-b$ is the product of the roots, we have that the two possibilities are $1 \cdot 9 \cdot (-10) = -90$ and $(-5)(-6)11 = 330$ . Adding the absolute values of these gives us $\boxed{420}$ | null | 420 |
a6066b7fc82ba1a949f59ca25a657bd3 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_6 | Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ . Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | The probability that he rolls a six twice when using the fair die is $\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$ . The probability that he rolls a six twice using the biased die is $\frac{2}{3}\times \frac{2}{3}=\frac{4}{9}=\frac{16}{36}$ . Given that Charles rolled two sixes, we can see that it is $16$ times more likely that he chose the second die. Therefore the probability that he is using the fair die is $\frac{1}{17}$ , and the probability that he is using the biased die is $\frac{16}{17}$ . The probability of rolling a third six is
\[\frac{1}{17}\times \frac{1}{6} + \frac{16}{17} \times \frac{2}{3} = \frac{1}{102}+\frac{32}{51}=\frac{65}{102}\] Therefore, our desired $p+q$ is $65+102= \boxed{167}$ | null | 167 |
a6066b7fc82ba1a949f59ca25a657bd3 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_6 | Charles has two six-sided die. One of the die is fair, and the other die is biased so that it comes up six with probability $\frac{2}{3}$ and each of the other five sides has probability $\frac{1}{15}$ . Charles chooses one of the two dice at random and rolls it three times. Given that the first two rolls are both sixes, the probability that the third roll will also be a six is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ | This is an incredibly simple problem if one is familiar with conditional probability (SO BE FAMILIAR WITH IT)!
The conditional probability that the third roll will be a six given that the first two rolls are sixes, is the conditional probability that Charles rolls three sixes given that his first two rolls are sixes. This is thus $\frac{\frac{1}{2}(\frac{2}{3})^3+\frac{1}{2}(\frac{1}{6})^3}{\frac{1}{2}(\frac{2}{3})^2+\frac{1}{2}(\frac{1}{6})^2}= \frac{65}{102}$ . The answer is therefore $65+102= \boxed{167}$ | null | 167 |
646ada4a6ad17a0e30bca8af6a1426ce | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_7 | Let $f(x)=(x^2+3x+2)^{\cos(\pi x)}$ . Find the sum of all positive integers $n$ for which \[\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.\] | First, let's split it into two cases to get rid of the absolute value sign
$\left |\sum_{k=1}^n\log_{10}f(k)\right|=1 \iff \sum_{k=1}^n\log_{10}f(k)=1,-1$
Now we simplify using product-sum logarithmic identites:
$\log_{10}{f(1)}+\log_{10}{f(2)}+...+\log_{10}{f(n)}=\log_{10}{f(1)\cdot f(2) \cdot ... \cdot f(n)}=1,-1$
Note that the exponent $\cos{\pi(x)}$ is either $-1$ if $x$ is odd or $1$ if $x$ is even.
Writing out the first terms we have
$\frac{1}{(2)(3)}(3)(4)\frac{1}{(4)(5)} \ldots$
This product clearly telescopes (i.e. most terms cancel) and equals either $10$ or $\frac{1}{10}$ . But the resulting term after telescoping depends on parity (odd/evenness), so we split it two cases, one where $n$ is odd and another where $n$ is even.
$\textbf{Case 1: Odd n}$
For odd $n$ , it telescopes to $\frac{1}{2(n+2)}$ where $n$ is clearly $3$
$\textbf{Case 2: Even n}$
For even $n$ , it telescopes to $\frac{n+2}{2}$ where $18$ is the only possible $n$ value. Thus the answer is $\boxed{021}$ | null | 021 |
646ada4a6ad17a0e30bca8af6a1426ce | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_7 | Let $f(x)=(x^2+3x+2)^{\cos(\pi x)}$ . Find the sum of all positive integers $n$ for which \[\left |\sum_{k=1}^n\log_{10}f(k)\right|=1.\] | Note that $\cos(\pi x)$ is $-1$ when $x$ is odd and $1$ when $x$ is even. Also note that $x^2+3x+2=(x+1)(x+2)$ for all $x$ . Therefore \[\log_{10}f(x)=\log_{10}(x+1)+\log_{10}(x+2)\ \ \ \text{if }x \text{ is even}\] \[\log_{10}f(x)=-\log_{10}(x+1)-\log_{10}(x+2)\ \ \ \text{if }x \text{ is odd}\] Because of this, $\sum_{k=1}^n\log_{10}f(k)$ is a telescoping series of logs, and we have \[\sum_{k=1}^n\log_{10}f(k)= \log_{10}(n+2)-\log_{10}2=\log_{10}\frac{n+2}{2}\ \ \ \text{if }n \text{ is even}\] \[\sum_{k=1}^n\log_{10}f(k)= -\log_{10}(n+2)-\log_{10}2=-\log_{10}2(n+2)\ \ \ \text{if }n \text{ is odd}\] Setting each of the above quantities to $1$ and $-1$ and solving for $n$ ,
we get possible values of $n=3$ and $n=18$ so our desired answer is $3+18=\boxed{021}$ | null | 021 |
c55f33206d90c44baeb03ad44886d8db | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8 | Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the form $\sqrt{m}-n$ , where $m$ and $n$ are positive integers. Find $m+n$ | Using the diagram above, let the radius of $D$ be $3r$ , and the radius of $E$ be $r$ . Then, $EF=r$ , and $CE=2-r$ , so the Pythagorean theorem in $\triangle CEF$ gives $CF=\sqrt{4-4r}$ . Also, $CD=CA-AD=2-3r$ , so \[DF=DC+CF=2-3r+\sqrt{4-4r}.\] Noting that $DE=4r$ , we can now use the Pythagorean theorem in $\triangle DEF$ to get \[(2-3r+\sqrt{4-4r})^2+r^2=16r^2.\]
Solving this quadratic is somewhat tedious, but the constant terms cancel, so the computation isn't terrible. Solving gives $3r=\sqrt{240}-14$ for a final answer of $\boxed{254}$ | null | 254 |
c55f33206d90c44baeb03ad44886d8db | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8 | Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the form $\sqrt{m}-n$ , where $m$ and $n$ are positive integers. Find $m+n$ | Consider a reflection of circle $E$ over diameter $\overline{AB}$ . By symmetry, we now have three circles that are pairwise externally tangent and all internally tangent to a large circle. The small circles have radii $r$ $r$ , and $3r$ , and the big circle has radius $2$
Descartes' Circle Theorem gives $\left(\frac{1}{r}+\frac{1}{r}+\frac{1}{3r}-\frac12\right)^2 = 2\left(\left(\frac{1}{r}\right)^2+\left(\frac{1}{r}\right)^2+\left(\frac{1}{3r}\right)^2+\left(-\frac12\right)^2\right)$
Note that the big circle has curvature $-\frac12$ because it is internally tangent.
Solving gives $3r=\sqrt{240}-14$ for a final answer of $\boxed{254}$ | null | 254 |
c55f33206d90c44baeb03ad44886d8db | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8 | Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the form $\sqrt{m}-n$ , where $m$ and $n$ are positive integers. Find $m+n$ | We use the notation of Solution 1 for triangle $\triangle DEC$ \[\sin \angle EDC = \frac {EF}{DE} = \frac {1}{4} \implies \cos \angle EDC = \frac {\sqrt{15}}{4}.\] We use Cosine Law for $\triangle DEC$ and get: \[(4r)^2 +(2 – 3r)^2 – 2 \cdot 4r \cdot (2 – 3r) \cdot \frac {\sqrt{15}}{4} = (2 – r)^2\] \[(24 + 6 \sqrt{15} ) r^2 = (8 + 4 \sqrt {15})r \implies 3r = 4 \sqrt{15} – 14 \implies \boxed{254}.\] vladimir.shelomovskii@gmail.com, vvsss | null | 254 |
c55f33206d90c44baeb03ad44886d8db | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_8 | Circle $C$ with radius 2 has diameter $\overline{AB}$ . Circle D is internally tangent to circle $C$ at $A$ . Circle $E$ is internally tangent to circle $C$ , externally tangent to circle $D$ , and tangent to $\overline{AB}$ . The radius of circle $D$ is three times the radius of circle $E$ , and can be written in the form $\sqrt{m}-n$ , where $m$ and $n$ are positive integers. Find $m+n$ | This problem can be very easily solved using Descartes' Circle Theorem. It states that if we have 4 circles that are all tangent with each other, $(k_1 + k_2 + k_3 + k_4)^{2} = 2(k_1^{2} + k_2^{2} + k_3^{2} + k_4^{2})$ , where $k_i$ is the curvature of circle $i$ , meaning $k_i = \dfrac{1}{r}$ . When three of the circles are internally tangent to the fourth one, the fourth circle has a negative curvature. Suppose we reflect Circle $E$ over $\overline{AB}$ . Now, we have our four circles to apply that theorem. First, lets scale our image down such that Circle $C$ has radius $1$ , for ease of computation. Let the radius of Circle $D$ be $r$ , so Circle $E$ has radius $\dfrac{r}{3}$ . Then, we have that $(-1 + \dfrac{1}{r} + \dfrac{3}{r} + \dfrac{3}{r})^{2} = 2(1 + \dfrac{1}{r^{2}} + \dfrac{9}{r^{2}} + \dfrac{9}{r^{2}})$ . This simplifies to $\dfrac{49}{r^{2}} - \dfrac{14}{r} + 1 = \dfrac{2r^{2} + 38}{r^{2}}$ . Multiplying both sides by $r^{2}$ , we get that $49 - 14r + r^{2} = 2r^{2} + 38$ , or $r^2 + 14r - 11 = 0$ . We get $r = -7 \pm 2\sqrt{15}$ , but we want the positive solution, which is $r = 2\sqrt{15} - 7$ . We have to rescale back up, so we get $r = 4\sqrt{15} - 14 = \sqrt{240} - 14$ , so we get that our answer is $240 + 14 = \boxed{254}$ .
~Puck_0 | null | 254 |
f242c719f277a2731f284413d8f248cd | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_9 | Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. | We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs. There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent. There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10 \cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have all $5$ adjacent, $10 \cdot 4$ or $40$ have $4$ adjacent, and $10 \cdot {5\choose 2}$ or $100$ have $3$ adjacent. With $6$ chairs in the subset, $10$ have all $6$ adjacent, $10(3)$ or $30$ have $5$ adjacent, $10 \cdot {4\choose2}$ or $60$ have $4$ adjacent, $\frac{10 \cdot 3}{2}$ or $15$ have $2$ groups of $3$ adjacent chairs, and $10 \cdot \left({5\choose2} - 3\right)$ or $70$ have $1$ group of $3$ adjacent chairs. All possible subsets with more than $6$ chairs have at least $1$ group of $3$ adjacent chairs, so we add ${10\choose7}$ or $120$ ${10\choose8}$ or $45$ ${10\choose9}$ or $10$ , and ${10\choose10}$ or $1.$ Adding, we get $10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = \boxed{581}.$ | null | 581 |
f242c719f277a2731f284413d8f248cd | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_9 | Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. | Starting with small cases, we see that four chairs give $4 + 1 = 5$ , five chairs give $5 + 5 + 1 = 11$ , and six chairs give $6 + 6 + 6 + 6 + 1 = 25.$ Thus, n chairs should give $n 2^{n-4} + 1$ , as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are $n 2^{n-3}$ ways to arrange $3$ adjacent chairs, but then we subtract $n 2^{n-4}$ ways to arrange $4.$ Finally, we add $1$ to account for the full subset of chairs. Thus, for $n = 10$ we get a first count of $641.$
However, we overcount cases in which there are two distinct groups of three or more chairs. We have $5$ cases for two groups of $3$ directly opposite each other, $5$ for two groups of four, $20$ for two groups of $3$ not symmetrically opposite, $20$ for a group of $3$ and a group of $4$ , and $10$ for a group of $3$ and a group of $5.$ Thus, we have $641 - 60 = \boxed{581}$ | null | 581 |
f242c719f277a2731f284413d8f248cd | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_9 | Ten chairs are arranged in a circle. Find the number of subsets of this set of chairs that contain at least three adjacent chairs. | It is possible to use recursion to count the complement. Number the chairs $1, 2, 3, ..., 10.$ If chair $1$ is not occupied, then we have a line of $9$ chairs such that there is no consecutive group of three. If chair $1$ is occupied, then we split into more cases. If chairs $2$ and $10$ are empty, then we have a line of $7.$ If chair $2$ is empty but chair $10$ is occupied, then we have a line of $6$ chairs (because chair $9$ cannot be occupied); this is similar to when chair $2$ is occupied and chair $10$ is empty. Finally, chairs $2$ and $10$ cannot be simultaneously occupied. Thus, we have reduced the problem down to computing $T_9 + T_7 + 2T_6$ , where $T_n$ counts the ways to select a subset of chairs $\textit{in a line}$ from a group of n chairs such that there is no group of $3$ chairs in a row.
Now, we notice that $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ (representing the cases when the first, second, and/or third chair is unoccupied). Also, $T_0 = 1, T_1 = 2, T_2 = 4, T_3 = 7$ , and hence $T_4 = 13, T_5 = 24, T_6 = 44, T_7 = 81, T_8 = 149, T_9 = 274$ . Now we know the complement is $274 + 81 + 88 = 443$ , and subtracting from $2^{10} = 1024$ gives $1024 - 443 = \boxed{581}$ | null | 581 |
d236e20e076cf82fe6ab48d30949e5ca | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_10 | Let $z$ be a complex number with $|z|=2014$ . Let $P$ be the polygon in the complex plane whose vertices are $z$ and every $w$ such that $\frac{1}{z+w}=\frac{1}{z}+\frac{1}{w}$ . Then the area enclosed by $P$ can be written in the form $n\sqrt{3}$ , where $n$ is an integer. Find the remainder when $n$ is divided by $1000$ | I find that generally, the trick to these kinds of AIME problems is to interpret the problem geometrically, and that is just what I did here. Looking at the initial equation, this seems like a difficult task, but rearranging yields a nicer equation: \[\frac1{z+w}=\frac1z+\frac1w\] \[\frac w{z+w}=\frac wz+1\] \[\frac w{z+w}=\frac{w+z}z\] \[\frac{w-0}{w-(-z)}=\frac{(-z)-w}{(-z)-0}\] We can interpret the difference of two complex numbers as a vector from one to the other, and we can interpret the quotient as a vector with an angle equal to the angle between the two vectors. Therefore, after labeling the complex numbers with $W$ $w$ ), $V$ $-z$ ), and $O$ $0$ ), we can interpret the above equation to mean that the $\angle OWV=\angle OVW$ , and hence triangle $OWV$ is isosceles, so length $OW$ $OV$ . Rearranging the equation \[\frac{w-0}{w-(-z)}=\frac{(-z)-w}{(-z)-0},\] we find that \[(w-0)((-z)-0)=-(w-(-z))^2.\] Taking the magnitude of both sides and using the fact that $OW=OV\implies |w-0|=|(-z)-0|$ , we find that \[|w-0|^2=|w-(-z)|^2,\] so length $OW=VW$ and triangle $OWV$ is equilateral. There are only two possible $W$ 's for which $OWV$ is equilateral, lying on either side of $OV$ . After drawing these points on the circle of radius 2014 centered at the origin, it is easy to see that $z$ and the two $w$ 's form an equilateral triangle (this can be verified by simple angle chasing). Drawing a perpendicular bisector of one of the sides and using 30-60-90 triangles shows that the side length of this triangle is $2014\sqrt3$ and hence its area is $\frac{\sqrt3(2014\sqrt3)^2}4=\boxed{147}\sqrt3+1000k\sqrt3,$ for some integer $k$ | null | 147 |
bacdb39177c11104da3165de02f4265e | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_11 | In $\triangle RED$ $\measuredangle DRE=75^{\circ}$ and $\measuredangle RED=45^{\circ}$ $RD=1$ . Let $M$ be the midpoint of segment $\overline{RD}$ . Point $C$ lies on side $\overline{ED}$ such that $\overline{RC}\perp\overline{EM}$ . Extend segment $\overline{DE}$ through $E$ to point $A$ such that $CA=AR$ . Then $AE=\frac{a-\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer. Find $a+b+c$ | Let $P$ be the foot of the perpendicular from $A$ to $\overline{CR}$ , so $\overline{AP}\parallel\overline{EM}$ . Since triangle $ARC$ is isosceles, $P$ is the midpoint of $\overline{CR}$ , and $\overline{PM}\parallel\overline{CD}$ . Thus, $APME$ is a parallelogram and $AE = PM = \frac{CD}{2}$ . We can then use coordinates. Let $O$ be the foot of altitude $RO$ and set $O$ as the origin. Now we notice special right triangles! In particular, $DO = \frac{1}{2}$ and $EO = RO = \frac{\sqrt{3}}{2}$ , so $D\left(\frac{1}{2}, 0\right)$ $E\left(-\frac{\sqrt{3}}{2}, 0\right)$ , and $R\left(0, \frac{\sqrt{3}}{2}\right).$ $M =$ midpoint $(D, R) = \left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$ and the slope of $ME = \frac{\frac{\sqrt{3}}{4}}{\frac{1}{4} + \frac{\sqrt{3}}{2}} = \frac{\sqrt{3}}{1 + 2\sqrt{3}}$ , so the slope of $RC = -\frac{1 + 2\sqrt{3}}{\sqrt{3}}.$ Instead of finding the equation of the line, we use the definition of slope: for every $CO = x$ to the left, we go $\frac{x(1 + 2\sqrt{3})}{\sqrt{3}} = \frac{\sqrt{3}}{2}$ up. Thus, $x = \frac{\frac{3}{2}}{1 + 2\sqrt{3}} = \frac{3}{4\sqrt{3} + 2} = \frac{3(4\sqrt{3} - 2)}{44} = \frac{6\sqrt{3} - 3}{22}.$ $DC = \frac{1}{2} - x = \frac{1}{2} - \frac{6\sqrt{3} - 3}{22} = \frac{14 - 6\sqrt{3}}{22}$ , and $AE = \frac{7 - \sqrt{27}}{22}$ , so the answer is $\boxed{056}$ | null | 056 |
5a2981deee447dff175747c6f2e2934d | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_12 | Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$ | WLOG, let C be the largest angle in the triangle.
As above, we can see that $\cos3A+\cos3B-\cos(3A+3B)=1$
Expanding, we get
$\cos3A+\cos3B-\cos3A\cos3B+\sin3A\sin3B=1$
$\cos3A\cos3B-\cos3A-\cos3B+1=\sin3A\sin3B$
$(\cos3A-1)(\cos3B-1)=\sin3A\sin3B$
CASE 1: If $\sin 3A = 0$ or $\sin 3B = 0$
This implies one or both of A or B are 60 or 120.
If one of A or B is 120, we have a contradiction, since C must be the largest angle.
Otherwise, if one of A or B is 60, WLOG, assume A = 60, we would have $\cos(3B) + \cos(3C) = 2$ , and thus, cos(3B) and cos(3C) both equal 1, implying $B = C = 120$ , a contradiction to the fact that the sum of the angles of a triangle must be 180 degrees.
CASE 2: If $\sin 3A \neq 0$ and $\sin 3B \neq 0$
$\frac{\cos3A-1}{\sin3A}\cdot\frac{\cos3B-1}{\sin3B}=1$
$\tan{\frac{3A}{2}}\tan{\frac{3B}{2}}=1$
Note that $\tan{x}=\frac{1}{\tan(90-x)}$ , or $\tan{x}\tan(90-x)=1$
Thus $\frac{3A}{2}+\frac{3B}{2}=90$ , or $A+B=60$
Now we know that $C=120$ , so we can just use the Law of Cosines to get $\boxed{399}$ | null | 399 |
5a2981deee447dff175747c6f2e2934d | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_12 | Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$ | \[\cos3A+\cos3B=1-\cos(3C)=1+\cos(3A+3B)\] \[2\cos\frac{3}{2}(A+B)\cos\frac{3}{2}(A-B)=2\cos^2\frac{3}{2}(A+B)\] If $\cos\frac{3}{2}(A+B) = 0$ , then $\frac{3}{2}(A+B)=90$ $A+B=60$ , so $C=120$ ; otherwise, \[2\cos\frac{3}{2}(A-B)=2cos\frac{3}{2}(A+B)\] \[\sin\frac{3}{2}A\sin\frac{3}{2}B=0\] so either $\sin\frac{3}{2}A=0$ or $\sin\frac{3}{2}B=0$ , i.e., either $A=120$ or $B=120$ . In all cases, one of the angles must be 120, which opposes the longest side. Final result follows. $\boxed{399}$ | null | 399 |
5a2981deee447dff175747c6f2e2934d | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_12 | Suppose that the angles of $\triangle ABC$ satisfy $\cos(3A)+\cos(3B)+\cos(3C)=1.$ Two sides of the triangle have lengths 10 and 13. There is a positive integer $m$ so that the maximum possible length for the remaining side of $\triangle ABC$ is $\sqrt{m}.$ Find $m.$ | Let $BC$ be the unknown side length. By Law of Cosines we have that $BC = \sqrt{269-260\cos{A}}$ . We notice that $\cos{A}$ should be negative to optimize $BC$ so $A$ is between $90$ and $180$ degrees. We also know that the value inside the square root is an integer $m$ , so $269-260\cos{A}$ should be an integer. We can then assume that $A$ is $120$ degrees so $\cos{A} = \frac{-1}{2}$ . We do this because $120$ degrees is a "common" value and it makes the value inside the square root an integer. Plugging this into $269-260\cos{A}$ for $m$ we get that it is $\boxed{399}$ | null | 399 |
7dc8eda4414d75f7f9938c79527d412c | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_13 | Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k<5$ , no collection of $k$ pairs made by the child contains the shoes from exactly $k$ of the adults is $\frac{m}{n}$ , where m and n are relatively prime positive integers. Find $m+n.$ | Label the left shoes be $L_1,\dots, L_{10}$ and the right shoes $R_1,\dots, R_{10}$ . Notice that there are $10!$ possible pairings.
Let a pairing be "bad" if it violates the stated condition. We would like a better condition to determine if a given pairing is bad.
Note that, in order to have a bad pairing, there must exist a collection of $k<5$ pairs that includes both the left and the right shoes of $k$ adults; in other words, it is bad if it is possible to pick $k$ pairs and properly redistribute all of its shoes to exactly $k$ people.
Thus, if a left shoe is a part of a bad collection, its corresponding right shoe must also be in the bad collection (and vice versa). To search for bad collections, we can start at an arbitrary right shoe (say $R_1$ ), check the left shoe it is paired with (say $L_i$ ), and from the previous observation, we know that $R_i$ must also be in the bad collection. Then we may check the left shoe paired with $R_i$ , find its counterpart, check its left pair, find its counterpart, etc. until we have found $L_1$ . We can imagine each right shoe "sending" us to another right shoe (via its paired left shoe) until we reach the starting right shoe, at which point we know that we have found a bad collection if we have done this less than $5$ times.
Effectively we have just traversed a cycle. (Note: This is the cycle notation of permutations.) The only condition for a bad pairing is that there is a cycle with length less than $5$ ; thus, we need to count pairings where every cycle has length at least $5$ . This is only possible if there is a single cycle of length $10$ or two cycles of length $5$
The first case yields $9!$ working pairings. The second case yields $\frac{{10\choose 5}}{2}\cdot{4!}^2=\frac{10!}{2 \cdot {5!}^2} \cdot {4!}^2$ pairings. Therefore, taking these cases out of a total of $10!$ , the probability is $\frac{1}{10}+\frac{1}{50} = \frac{3}{25}$ , for an answer of $\boxed{028}$ | null | 028 |
6abf82269fc3be9b04be60f83d4c3a19 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_14 | In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$ . Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$ $\angle{BAD}=\angle{CAD}$ , and $BM=CM$ . Point $N$ is the midpoint of the segment $HM$ , and point $P$ is on ray $AD$ such that $PN\perp{BC}$ . Then $AP^2=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let us just drop the perpendicular from $B$ to $AC$ and label the point of intersection $O$ . We will use this point later in the problem.
As we can see,
$M$ is the midpoint of $BC$ and $N$ is the midpoint of $HM$
$AHC$ is a $45-45-90$ triangle, so $\angle{HAB}=15^\circ$
$AHD$ is $30-60-90$ triangle.
$AH$ and $PN$ are parallel lines so $PND$ is $30-60-90$ triangle also.
Then if we use those informations we get $AD=2HD$ and
$PD=2ND$ and $AP=AD-PD=2HD-2ND=2HN$ or $AP=2HN=HM$
Now we know that $HM=AP$ , we can find for $HM$ which is simpler to find.
We can use point $B$ to split it up as $HM=HB+BM$
We can chase those lengths and we would get
$AB=10$ , so $OB=5$ , so $BC=5\sqrt{2}$ , so $BM=\dfrac{1}{2} \cdot BC=\dfrac{5\sqrt{2}}{2}$
We can also use Law of Sines:
\[\frac{BC}{AB}=\frac{\sin\angle A}{\sin\angle C}\] \[\frac{BC}{10}=\frac{\frac{1}{2}}{\frac{\sqrt{2}}{2}}\implies BC=5\sqrt{2}\]
Then using right triangle $AHB$ , we have $HB=10 \sin 15^\circ$
So $HB=10 \sin 15^\circ=\dfrac{5(\sqrt{6}-\sqrt{2})}{2}$
And we know that $AP = HM = HB + BM = \frac{5(\sqrt6-\sqrt2)}{2} + \frac{5\sqrt2}{2} = \frac{5\sqrt6}{2}$
Finally if we calculate $(AP)^2$
$(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}$ . So our final answer is $75+2=77$
$m+n=\boxed{077}$ | null | 077 |
6abf82269fc3be9b04be60f83d4c3a19 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_14 | In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$ . Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$ $\angle{BAD}=\angle{CAD}$ , and $BM=CM$ . Point $N$ is the midpoint of the segment $HM$ , and point $P$ is on ray $AD$ such that $PN\perp{BC}$ . Then $AP^2=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Break our diagram into 2 special right triangle by dropping an altitude from $B$ to $AC$ we then get that \[AC=5+5\sqrt{3}, BC=5\sqrt{2}.\] Since $\triangle{HCA}$ is a 45-45-90,
\[HC=\frac{5\sqrt2+5\sqrt6}{2}\] $MC=\frac{BC}{2},$ \[HM=\frac{5\sqrt6}{2}\] \[HN=\frac{5\sqrt6}{4}\] We know that $\triangle{AHD}\simeq \triangle{PND}$ and are 30-60-90.
Thus, \[AP=2 \cdot HN=\frac{5\sqrt6}{2}.\]
$(AP)^2=\dfrac{150}{4}=\dfrac{75}{2}$ . So our final answer is $75+2=\boxed{077}$ | null | 077 |
6abf82269fc3be9b04be60f83d4c3a19 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_14 | In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$ . Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$ $\angle{BAD}=\angle{CAD}$ , and $BM=CM$ . Point $N$ is the midpoint of the segment $HM$ , and point $P$ is on ray $AD$ such that $PN\perp{BC}$ . Then $AP^2=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -8.455641974276588, xmax = 26.731282460265, ymin = -10.92318356252699, ymax = 9.023689834456471; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819)--cycle, linewidth(2) + rvwvcq); /* draw figures */ draw((-1.4934334172297545,2.6953043701763835)--(1.1286284157632023,-6.954814372303504), linewidth(2) + wrwrwr); draw((xmin, -0.9930079421029264*xmin + 1.2123131258653241)--(xmax, -0.9930079421029264*xmax + 1.2123131258653241), linewidth(2) + wrwrwr); /* line */ draw((xmin, 0.0035082940460819836*xmin-6.958773932654766)--(xmax, 0.0035082940460819836*xmax-6.958773932654766), linewidth(2) + wrwrwr); /* line */ draw((xmin, -285.03882139434313*xmin-422.9911967079192)--(xmax, -285.03882139434313*xmax-422.9911967079192), linewidth(2) + wrwrwr); /* line */ draw((xmin, -1.7181023895538718*xmin + 0.12943284739433739)--(xmax, -1.7181023895538718*xmax + 0.12943284739433739), linewidth(2) + wrwrwr); /* line */ draw(circle((4.642656870506668,-0.8187240845670819), 7.071067811865476), linewidth(2) + wrwrwr); draw((xmin, -285.0388213943529*xmin + 1322.5187184230485)--(xmax, -285.0388213943529*xmax + 1322.5187184230485), linewidth(2) + wrwrwr); /* line */ draw((-1.4934334172297545,2.6953043701763835)--(4.617849638067675,6.252300211899359), linewidth(2) + wrwrwr); draw((4.617849638067675,6.252300211899359)--(-1.459546107520503,-6.96389444957376), linewidth(2) + wrwrwr); draw(circle((-0.18240250073327363,-2.12975500106356), 5), linewidth(2) + wrwrwr); draw((xmin, -285.0388213943432*xmin + 449.7637608575419)--(xmax, -285.0388213943432*xmax + 449.7637608575419), linewidth(2) + wrwrwr); /* line */ draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + wrwrwr); draw((-1.4934334172297545,2.6953043701763835)--(4.642656870506668,-0.8187240845670819), linewidth(2) + wrwrwr); draw((-1.4934334172297545,2.6953043701763835)--(-1.459546107520503,-6.96389444957376), linewidth(2) + rvwvcq); draw((-1.459546107520503,-6.96389444957376)--(1.1286284157632023,-6.954814372303504), linewidth(2) + rvwvcq); draw((1.1286284157632023,-6.954814372303504)--(4.651736947776926,-3.406898607850789), linewidth(2) + rvwvcq); draw((4.651736947776926,-3.406898607850789)--(4.642656870506668,-0.8187240845670819), linewidth(2) + rvwvcq); draw((4.642656870506668,-0.8187240845670819)--(-1.4934334172297545,2.6953043701763835), linewidth(2) + rvwvcq); /* dots and labels */ dot((-1.4934334172297545,2.6953043701763835),dotstyle); label("$A$", (-1.3954084351380491,2.9230996889873015), NE * labelscalefactor); dot((1.1286284157632023,-6.954814372303504),dotstyle); label("$B$", (1.2093379191072373,-6.719031552166216), NE * labelscalefactor); dot((8.199652712229643,-6.930007139864511),linewidth(4pt) + dotstyle); label("$C$", (8.292420110475998,-6.741880204396438), NE * labelscalefactor); dot((-1.459546107520503,-6.96389444957376),linewidth(4pt) + dotstyle); label("$H$", (-1.3725597829078273,-6.787577508856881), NE * labelscalefactor); dot((-1.4686261847907602,-4.375719926290057),linewidth(4pt) + dotstyle); label("$E$", (-1.3725597829078273,-4.182831154611618), NE * labelscalefactor); dot((4.617849638067675,6.252300211899359),linewidth(4pt) + dotstyle); label("$L$", (4.705181710331174,6.441792132441429), NE * labelscalefactor); dot((4.642656870506668,-0.8187240845670819),linewidth(4pt) + dotstyle); label("$O$", (4.728030362561396,-0.6412900589272691), NE * labelscalefactor); dot((4.117194931218359,-6.944329602191013),linewidth(4pt) + dotstyle); label("$D$", (4.2025113612662945,-6.764728856626659), NE * labelscalefactor); dot((4.6674641029456625,-7.889748381033524),linewidth(4pt) + dotstyle); label("$F$", (4.750879014791618,-7.701523598065745), NE * labelscalefactor); dot((4.651736947776926,-3.406898607850789),linewidth(4pt) + dotstyle); label("$G$", (4.750879014791618,-3.2231877609423107), NE * labelscalefactor); dot((1.5791517652735851,-0.3557971188372022),linewidth(4pt) + dotstyle); label("$K$", (1.6663109637116735,-0.18431701432283698), NE * labelscalefactor); dot((1.5870153428579534,-2.5972220054285695),linewidth(4pt) + dotstyle); label("$P$", (1.6891596159418953,-2.4234849328845542), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy] Draw the $45-45-90 \triangle AHC$ . Now, take the perpendicular bisector of $BC$ to intersect the circumcircle of $\triangle ABC$ and $AC$ at $F, L, G$ as shown, and denote $O$ to be the circumcenter of $\triangle ABC$ . It is not difficult to see by angle chasing that $AHBGO$ is cyclic, namely with diameter $AB$ . Then, by symmetry, $EH = HB$ and as $HB, OG$ are both subtended by equal arcs they are equal. Hence, $EH = GO$ . Now, draw line $HL$ and intersect it at $AC$ at point $K$ in the diagram. It is not hard to use angle chase to arrive at $AEOL$ a parallelogram, and from our length condition derived earlier, $AL \parallel HG$ . From here, it is clear that $AK = KG$ ; that is, $P$ is just the intersection of the perpendicular from $K$ down to $BC$ and $AD$ ! After this point, note that $AP = PF$ . It is easily derived that the circumradius of $\triangle ABC$ is $\frac{10}{\sqrt{2}}$ . Now, $APO$ is a $30-60-90$ triangle, and from here it is easy to arrive at the final answer of $\boxed{077}$ . ~awang11's sol | null | 077 |
6abf82269fc3be9b04be60f83d4c3a19 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_14 | In $\triangle{ABC}, AB=10, \angle{A}=30^\circ$ , and $\angle{C=45^\circ}$ . Let $H, D,$ and $M$ be points on the line $BC$ such that $AH\perp{BC}$ $\angle{BAD}=\angle{CAD}$ , and $BM=CM$ . Point $N$ is the midpoint of the segment $HM$ , and point $P$ is on ray $AD$ such that $PN\perp{BC}$ . Then $AP^2=\dfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ | Let $BO \perp AC, O \in AC.$
Let $ME \perp BC, E \in AD.$
$MB = MC, \angle C = 45^\circ \implies$ points $M, E, O$ are collinear.
$HN = NM, AH||NP||ME \implies AP = PE.$
In $\triangle ABO \hspace{10mm} \angle A = 30^\circ \implies AO = AB \cos 60^\circ = 5 \sqrt{3}.$
In $\triangle AEO \hspace{10mm} \angle A = 15^\circ, \angle O = 90^\circ + 45^\circ = 135^\circ \implies$ \[\angle AEO = 30^\circ \implies\] \[AE = AO \frac {\sin 135^\circ}{\sin 30^\circ} = 5 \sqrt{3} \cdot \sqrt{2} = 5 \sqrt{6} \implies\] \[AP = 5 \sqrt {\frac {3}{2}} \implies AP^2 = \frac {75}{2} \implies \boxed{077}.\] vladimir.shelomovskii@gmail.com, vvsss | null | 077 |
bde8d4828f733ce5dfa7e0c55c8773e6 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_15 | For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the smallest positive integer $t$ such that $x_t=2090.$ | Note that for any $x_i$ , for any prime $p$ $p^2 \nmid x_i$ . This provides motivation to translate $x_i$ into a binary sequence $y_i$
Let the prime factorization of $x_i$ be written as $p_{a_1} \cdot p_{a_2} \cdot p_{a_3} \cdots$ , where $p_i$ is the $i$ th prime number. Then, for every $p_{a_k}$ in the prime factorization of $x_i$ , place a $1$ in the $a_k$ th digit of $y_i$ . This will result in the conversion $x_1 = 2, x_{2} = 3, x_{3} = 2 * 3 = 6, \cdots$
Multiplication for the sequence $x_i$ will translate to addition for the sequence $y_i$ . Thus, we see that $x_{n+1}X(x_n) = x_np(x_n)$ translates into $y_{n+1} = y_n+1$ . Since $x_0=1$ , and $y_0=0$ $x_i$ corresponds to $y_i$ , which is $i$ in binary. Since $x_{10010101_2} = 2 \cdot 5 \cdot 11 \cdot 19 = 2090$ $t = 10010101_2$ $\boxed{149}$ | null | 149 |
bde8d4828f733ce5dfa7e0c55c8773e6 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_15 | For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the smallest positive integer $t$ such that $x_t=2090.$ | We go through the terms and look for a pattern. We find that
$x_0 = 1$ $x_8 = 7$
$x_1 = 2$ $x_9 = 14$
$x_2 = 3$ $x_{10} = 21$
$x_3 = 6$ $x_{11} = 42$
$x_4 = 5$ $x_{12} = 35$
$x_5 = 10$ $x_{13} = 70$
$x_6 = 15$ $x_{14} = 105$
$x_7 = 30$ $x_{15} = 210$
Commit to the bash. Eventually, you will receive that $x_{149} = 2090$ , so $\boxed{149}$ is the answer. Trust me, this is worth the 10 index points. | null | 149 |
bde8d4828f733ce5dfa7e0c55c8773e6 | https://artofproblemsolving.com/wiki/index.php/2014_AIME_II_Problems/Problem_15 | For any integer $k\geq 1$ , let $p(k)$ be the smallest prime which does not divide $k.$ Define the integer function $X(k)$ to be the product of all primes less than $p(k)$ if $p(k)>2$ , and $X(k)=1$ if $p(k)=2.$ Let $\{x_n\}$ be the sequence defined by $x_0=1$ , and $x_{n+1}X(x_n)=x_np(x_n)$ for $n\geq 0.$ Find the smallest positive integer $t$ such that $x_t=2090.$ | Let $P_k$ denote the $k$ th prime.
Lemma: $x_{n+2^{k-1}} = P_k \cdot x_{n}$ for all $0 \leq n \leq 2^{k-1} - 1.$
$\mathbf{\mathrm{Proof:}}$
We can prove this using induction. Assume that $x_{2^{k-1}-1} = \prod_{j=1}^{k-1} P_j.$ Then, using the given recursion $x_{k+1} = \frac{x_np(x_n)}{X(x_n)}$ , we would “start fresh” for $x_{2^{k-1}} = P_k.$ It is then easy to see that then $\frac{x_n}{P_k}$ just cycles through the previous $x_{2^{k-1}}$ terms of $\{ x_n \},$ since the recursion process is the same and $P_k$ being a factor of $x_n$ is not affected until $n = 2 \cdot {2^{k-1}} = 2^k,$ when given our assumption $x_{2^k - 1} = \prod_{j=1}^{k} P_j$ and $n = 2^k$ is now the least $n$ such that \[P_{k+1} = p(x_{2^{n-1}}),\] in which $P_a = p(x_n)$ where $a > k$ is the only way that the aforementioned cycle would be affected. Specifically, by cancellation according to our recursion, $x_{2^k} = P_{k+1},$ and the values of $x_n$ just starts cycling through the previous $x_{2^k}$ terms again until $x_{2^{k+1}}$ when a new prime shows up in the prime factorization of $x_n,$ when it starts cycling again, and so on. Using our base cases of $x_0$ and $x_1,$ our induction is complete.
Now, it is easy to see that $2090 = 2 \cdot 5 \cdot 11 \cdot 19 = P_1 \cdot P_3 \cdot P_5 \cdot P_8,$ and by Lemma #1, the least positive integer $n$ such that $19 | x_n$ is $2^7.$ By repeatedly applying our obtained recursion from Lemma #1, it is easy to see that our answer is just $2^7 + 2^4 + 2^2 + 2^0,$ or $10010101_2 = \boxed{149}.$ | null | 149 |
7059ae2b6e1fdcf3de6e9ad489fedfc2 | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_1 | The AIME Triathlon consists of a half-mile swim, a 30-mile bicycle ride, and an eight-mile run. Tom swims, bicycles, and runs at constant rates. He runs fives times as fast as he swims, and he bicycles twice as fast as he runs. Tom completes the AIME Triathlon in four and a quarter hours. How many minutes does he spend bicycling? | Let $r$ represent the rate Tom swims in miles per minute. Then we have
$\frac{1/2}{r} + \frac{8}{5r} + \frac{30}{10r} = 255$
Solving for $r$ , we find $r = 1/50$ , so the time Tom spends biking is $\frac{30}{(10)(1/50)} = \boxed{150}$ minutes. | null | 150 |
24109d5876b94822acd009c30413435b | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_2 | Find the number of five-digit positive integers, $n$ , that satisfy the following conditions: | The number takes a form of $5\text{x,y,z}5$ , in which $5|(x+y+z)$ . Let $x$ and $y$ be arbitrary digits. For each pair of $x,y$ , there are exactly two values of $z$ that satisfy the condition of $5|(x+y+z)$ . Therefore, the answer is $10\times10\times2=\boxed{200}$ | null | 200 |
749be077979b0f39ac838b83e0f388fc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_3 | Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$ | It's important to note that $\dfrac{AE}{EB} + \dfrac{EB}{AE}$ is equivalent to $\dfrac{AE^2 + EB^2}{(AE)(EB)}$
We define $a$ as the length of the side of larger inner square, which is also $EB$ $b$ as the length of the side of the smaller inner square which is also $AE$ , and $s$ as the side length of $ABCD$ . Since we are given that the sum of the areas of the two squares is $\frac{9}{10}$ of the the area of ABCD, we can represent that as $a^2 + b^2 = \frac{9s^2}{10}$ . The sum of the two nonsquare rectangles can then be represented as $2ab = \frac{s^2}{10}$
Looking back at what we need to find, we can represent $\dfrac{AE^2 + EB^2}{(AE)(EB)}$ as $\dfrac{a^2 + b^2}{ab}$ . We have the numerator, and dividing $\frac{s^2}{10}$ by two gives us the denominator $\frac{s^2}{20}$ . Dividing $\dfrac{\frac{9s^2}{10}}{\frac{s^2}{20}}$ gives us an answer of $\boxed{018}$ | null | 018 |
749be077979b0f39ac838b83e0f388fc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_3 | Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$ | Let the side of the square be $1$ . Therefore the area of the square is also $1$ .
We label $AE$ as $a$ and $EB$ as $b$ . Notice that what we need to find is equivalent to: $\frac{a^2+b^2}{ab}$ .
Since the sum of the two squares ( $a^2+b^2$ ) is $\frac{9}{10}$ (as stated in the problem) the area of the whole square, it is clear that the
sum of the two rectangles is $1-\frac{9}{10} \implies \frac{1}{10}$ . Since these two rectangles are congruent, they
each have area: $\frac{1}{20}$ . Also note that the area of this is $ab$ . Plugging this into our equation we get:
$\frac{\frac{9}{10}}{\frac{1}{20}} \implies \boxed{018}$ | null | 018 |
749be077979b0f39ac838b83e0f388fc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_3 | Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$ | Let $AE$ be $x$ , and $EB$ be $1$ . Then we are looking for the value $x+\frac{1}{x}$ . The areas of the smaller squares add up to $9/10$ of the area of the large square, $(x+1)^2$ . Cross multiplying and simplifying we get $x^2-18x+1=0$ . Rearranging, we get $x+\frac{1}{x}=\boxed{018}$ | null | 018 |
749be077979b0f39ac838b83e0f388fc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_3 | Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$ | As before, $\dfrac{AE}{EB} + \dfrac{EB}{AE}$ is equivalent to $\dfrac{AE^2 + EB^2}{(AE)(EB)}$ . Let $x$ represent the value of $AE=CF$ . Since $EB=FB=1-x,$ the area of the two rectangles is $2x(1-x)=-2x^2+2x=\frac1{10}$ . Adding $2x^2-2x$ to both sides and dividing by $2$ gives $x^2-x+\frac1{20}=0.$ Note that the two possible values of $x$ in the quadratic both sum to $1,$ like how $AE$ and $EB$ does. Therefore, $EB$ must be the other root of the quadratic that $AE$ isn't. Applying Vietas and manipulating the numerator, we get $\frac{x_1^2+x_2^2}{x_1x_2}=\frac{(x_1+x_2)^2-2x_1x_2}{\frac{1}{20}}=\frac{1^2-\frac1{10}}{\frac1{20}}=\frac{\frac9{10}}{\frac{1}{20}}=\boxed{018}$ | null | 018 |
749be077979b0f39ac838b83e0f388fc | https://artofproblemsolving.com/wiki/index.php/2013_AIME_I_Problems/Problem_3 | Let $ABCD$ be a square, and let $E$ and $F$ be points on $\overline{AB}$ and $\overline{BC},$ respectively. The line through $E$ parallel to $\overline{BC}$ and the line through $F$ parallel to $\overline{AB}$ divide $ABCD$ into two squares and two nonsquare rectangles. The sum of the areas of the two squares is $\frac{9}{10}$ of the area of square $ABCD.$ Find $\frac{AE}{EB} + \frac{EB}{AE}.$ | Let $AE = x$ and $BE = y$ . From this, we get $AB = x + y$ . The problem is asking for $\frac{x}{y} + \frac{y}{x}$ , which can be rearranged to give $\frac{x^2 + y^2}{xy}$ . The problem tells us that $x^2 + y^2 = \frac{9(x+y)^2}{10}$ . We simplify to get $x^2 + y^2 = 18xy$ . Finally, we divide both sides by $xy$ to get $\frac{x^2 + y^2}{xy} = \boxed{018}$ . - Spacesam | null | 018 |