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Difference between revisions of "2007 AIME II Problems/Problem 3"

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=== Solution 2 ===
 
=== Solution 2 ===
  
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Extend <math>\overline{AE}, \overline{DF}</math> and <math>\overline{BE}, \overline{CF}</math> to their points of intersection. Since <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two new triangles are also congruent to these two (use [[ASA]], as we know all the sides are <math>13</math> and the angles are mostly complementary). Thus, we create a [[square]] with sides <math>5 + 12 = 17</math>.
+
Extend <math>\overline{AE}, \overline{DF}</math> and <math>\overline{BE}, \overline{CF}</math> to their points of intersection. Since <math>\triangle ABE \cong \triangle CDF</math> and are both <math>5-12-13</math> [[right triangle]]s, we can come to the conclusion that the two new triangles are also congruent to these two (use [https://artofproblemsolving.com/wiki/index.php/Congruent_(geometry)#ASA_Congruence ASA], as we know all the sides are <math>13</math> and the angles are mostly complementary). Thus, we create a [[square]] with sides <math>5 + 12 = 17</math>.
  
 
<asy>unitsize(0.25 cm);
 
<asy>unitsize(0.25 cm);
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===Solution 5 (Ptolemy's Theorem)===
 
===Solution 5 (Ptolemy's Theorem)===
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Drawing <math>EF</math>, it clearly passes through the center of <math>ABCD</math>. Letting this point be <math>P</math>, we note that <math>AEBP</math> and <math>CFDP</math> are congruent cyclic quadrilaterals, and that <math>AP=BP=CP=DP=\frac{13}{\sqrt{2}}.</math> Now, from Ptolemy's, <math>13\cdot EP=\frac{13}{\sqrt{2}}(12+5)\implies EP+\frac{17\sqrt{2}}{2}</math>. Since <math>EF=EP+FP=2\cdot EP</math>, the answer is <math>(17\sqrt{2})^2=\boxed{578}.</math>
+
Drawing <math>EF</math>, it clearly passes through the center of <math>ABCD</math>. Letting this point be <math>P</math>, we note that <math>AEBP</math> and <math>CFDP</math> are congruent cyclic quadrilaterals, and that <math>AP=BP=CP=DP=\frac{13}{\sqrt{2}}.</math> Now, from Ptolemy's, <math>13\cdot EP=\frac{13}{\sqrt{2}}(12+5)\implies EP=\frac{17\sqrt{2}}{2}</math>. Since <math>EF=EP+FP=2\cdot EP</math>, the answer is <math>(17\sqrt{2})^2=\boxed{578}.</math>
  
 
===Solution 6===
 
===Solution 6===
 
Coordinate bash
 
Coordinate bash
  
 +
=== Solution 7 (Trig Bash) ===
  
−
== Solution 7 (Trig Bash) ==
+
We first see that the whole figure is symmetrical and reflections across the center that we will denote as <math>O</math> bring each half of the figure to the other half. Thus we consider a single part of the figure, namely <math>EO.</math>
  
−
We first See that the whole figure is symmetrical and reflections across the center that we will denote as <math>O</math> bring each half of the figure to the other half. Thus we consider a single part of the figure, namely <math>EO.</math>
+
First note that <math>\angle BAO = 45^{\circ}</math> since <math>O</math> is the center of square <math>ABCD.</math> Also note that <math>\angle EAB = \arccos{\left(\frac{12}{13}\right)}</math> or <math>\arcsin{\left(\frac{5}{13}\right)}.</math> Finally, we know that <math>AO =\frac{13\sqrt{2}}{2}.</math> Now we apply laws of cosines on <math>\bigtriangleup AEO.</math>  
  
−
First note that <math>\angle BAO = 45^{\circ}</math> since <math>O</math> is the center of square <math>ABCD.</math> Also note that <math>\angle EAB = \arccos{\left(\frac{12}{13}\right)}</math> or <math>\arcsin{\left(\frac{5}{13}\right)}.</math> Finally, we know that <math>AO =\frac{13\sqrt{2}}{2}.</math> Now we apply laws of cosines on <math>\bigtriangleup AEO.</math>  
+
We have <math>EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}.</math> We know that <math>\angle EAO = 45^{\circ} + \arccos{\left(\frac{12}{13}\right)}.</math> Thus we have <math>\cos{\angle EAO} = \cos\left(45^{\circ} + \arccos{\left(\frac{12}{13}\right)}\right)</math> which applying the cosine sum identity yields <math>\cos{45^{\circ}}\cos{\arccos\frac{12}{13}} - \sin{45^{\circ}}\sin\arcsin{\frac{5}{12}} =\frac{12\sqrt{2}}{26} -\frac{5\sqrt{2}}{26} =\frac{7\sqrt{2}}{26}.</math>  
  
−
We have <math>EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot\frac{13\sqrt{2}}{2} \cdot \cos{\angle EAO}.</math> We know that <math>\angle EAO = 45 + \arccos{\left(\frac{12}{13}\right)}</math>
+
Note that we are looking for <math>4EO^2</math> so we multiply <math>EO^2 = 12^2 + \left(\frac{13\sqrt{2}}{2}\right)^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}</math> by <math>4</math> obtaining <math>4EO^2 = 576 + 338 - 8 \cdot\left(\frac{13\sqrt{2}}{2}\right) \cdot 12 \cdot\frac{7\sqrt{2}}{26} = 576 + 338 - 4  \cdot 12 \cdot  7 = \boxed{578}.</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 23:05, 7 July 2023

Problem

Square $ABCD$ has side length $13$, and points $E$ and $F$ are exterior to the square such that $BE=DF=5$ and $AE=CF=12$. Find $EF^{2}$. [asy]unitsize(0.2 cm); pair A, B, C, D, E, F; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); draw(A--B--C--D--cycle); draw(A--E--B); draw(C--F--D); dot("$A$", A, W); dot("$B$", B, dir(0)); dot("$C$", C, dir(0)); dot("$D$", D, W); dot("$E$", E, N); dot("$F$", F, S);[/asy]

Solution

Solution 1

Let $\angle FCD = \alpha$, so that $FB = \sqrt{12^2 + 13^2 + 2\cdot12\cdot13\sin(\alpha)} = \sqrt{433}$. By the diagonal, $DB = 13\sqrt{2}, DB^2 = 338$.

The sum of the squares of the sides of a parallelogram is the sum of the squares of the diagonals. \[EF^2 = 2\cdot(5^2 + 433) - 338 = 578.\]

Solution 2

Extend $\overline{AE}, \overline{DF}$ and $\overline{BE}, \overline{CF}$ to their points of intersection. Since $\triangle ABE \cong \triangle CDF$ and are both $5-12-13$ right triangles, we can come to the conclusion that the two new triangles are also congruent to these two (use ASA, as we know all the sides are $13$ and the angles are mostly complementary). Thus, we create a square with sides $5 + 12 = 17$.

[asy]unitsize(0.25 cm); pair A, B, C, D, E, F, G, H; A = (0,13); B = (13,13); C = (13,0); D = (0,0); E = A + (12*12/13,5*12/13); F = D + (5*5/13,-5*12/13); G = rotate(90,(A + C)/2)*(E); H = rotate(90,(A + C)/2)*(F); draw(A--B--C--D--cycle); draw(E--G--F--H--cycle); dot("$A$", A, N); dot("$B$", B, dir(0)); dot("$C$", C, S); dot("$D$", D, W); dot("$E$", E, N); dot("$F$", F, S); dot("$G$", G, W); dot("$H$", H, dir(0));[/asy]

$\overline{EF}$ is the diagonal of the square, with length $17\sqrt{2}$; the answer is $EF^2 = (17\sqrt{2})^2 = 578$.

Solution 3

A slightly more analytic/brute-force approach:

AIME II prob10 bruteforce.PNG

Drop perpendiculars from $E$ and $F$ to $I$ and $J$, respectively; construct right triangle $EKF$ with right angle at K and $EK || BC$. Since $2[CDF]=DF*CF=CD*JF$, we have $JF=5\times12/13 = \frac{60}{13}$. Similarly, $EI=\frac{60}{13}$. Since $\triangle DJF \sim \triangle DFC$, we have $DJ=\frac{5JF}{12}=\frac{25}{13}$.

Now, we see that $FK=DC-(DJ+IB)=DC-2DJ=13-\frac{50}{13}=\frac{119}{13}$. Also, $EK=BC+(JF+IE)=BC+2JF=13+\frac{120}{13}=\frac{289}{13}$. By the Pythagorean Theorem, we have $EF=\sqrt{\left(\frac{289}{13}\right)^2+\left(\frac{119}{13} \right)^2}=\frac{\sqrt{(17^2)(17^2+7^2)}}{13}$$=\frac{17\sqrt{338}}{13}=\frac{17(13\sqrt{2})}{13}=17\sqrt{2}$. Therefore, $EF^2=(17\sqrt{2})^2=578$.

Solution 4

Based on the symmetry, we know that $F$ is a reflection of $E$ across the center of the square, which we will denote as $O$. Since $\angle BEA$ and $\angle AOB$ are right, we can conclude that figure $AOBE$ is a cyclic quadrilateral. Pythagorean Theorem yields that $BO=AO=\frac{13\sqrt{2}}{2}$. Now, using Ptolemy's Theorem, we get that \[AO\cdot BE + BO\cdot AE = AB\cdot AO\] \[\frac{13\sqrt{2}}{2}\cdot 5+\frac{13\sqrt{2}}{2}\cdot 12 = 13\cdot OE\] \[OE=\frac{17\sqrt{2}}{2}\] Now, since we stated in the first step that $F$ is a reflection of $E$ across $O$, we can say that $EF=2EO=17\sqrt{2}$. This gives that \[EF^2=(17\sqrt{2})^2=578\] AWD with this bash solution

Solution 5 (Ptolemy's Theorem)

Drawing $EF$, it clearly passes through the center of $ABCD$. Letting this point be $P$, we note that $AEBP$ and $CFDP$ are congruent cyclic quadrilaterals, and that $AP=BP=CP=DP=\frac{13}{\sqrt{2}}.$ Now, from Ptolemy's, $13\cdot EP=\frac{13}{\sqrt{2}}(12+5)\implies EP=\frac{17\sqrt{2}}{2}$. Since $EF=EP+FP=2\cdot EP$, the answer is $(17\sqrt{2})^2=\boxed{578}.$

Solution 6

Coordinate bash

Solution 7 (Trig Bash)

We first see that the whole figure is symmetrical and reflections across the center that we will denote as $O$ bring each half of the figure to the other half. Thus we consider a single part of the figure, namely $EO.$

First note that $\angle BAO = 45^{\circ}$ since $O$ is the center of square $ABCD.$ Also note that $\angle EAB = \arccos{\left(\frac{12}{13}\right)}$ or $\arcsin{\left(\frac{5}{13}\right)}.$ Finally, we know that $AO =\frac{13\sqrt{2}}{2}.$ Now we apply laws of cosines on $\bigtriangleup AEO.$

We have $EO^2 = 12^2 + (\frac{13\sqrt{2}}{2})^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}.$ We know that $\angle EAO = 45^{\circ} + \arccos{\left(\frac{12}{13}\right)}.$ Thus we have $\cos{\angle EAO} = \cos\left(45^{\circ} + \arccos{\left(\frac{12}{13}\right)}\right)$ which applying the cosine sum identity yields $\cos{45^{\circ}}\cos{\arccos\frac{12}{13}} - \sin{45^{\circ}}\sin\arcsin{\frac{5}{12}} =\frac{12\sqrt{2}}{26} -\frac{5\sqrt{2}}{26} =\frac{7\sqrt{2}}{26}.$

Note that we are looking for $4EO^2$ so we multiply $EO^2 = 12^2 + \left(\frac{13\sqrt{2}}{2}\right)^2 - 2 \cdot 12 \cdot \left(\frac{13\sqrt{2}}{2}\right) \cdot \cos{\angle EAO}$ by $4$ obtaining $4EO^2 = 576 + 338 - 8 \cdot\left(\frac{13\sqrt{2}}{2}\right) \cdot 12 \cdot\frac{7\sqrt{2}}{26} = 576 + 338 - 4 \cdot 12 \cdot 7 = \boxed{578}.$

See also

2007 AIME II (Problems • Answer Key • Resources)
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