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

(Solution 7 (Trig Bash))
 
(24 intermediate revisions by 8 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
[[Square]] <math>ABCD</math> has side length <math>13</math>, and [[point]]s <math>E</math> and <math>F</math> are exterior to the square such that <math>BE=DF=5</math> and <math>AE=CF=12</math>. Find <math>\displaystyle EF^{2}</math>.
+
[[Square]] <math>ABCD</math> has side length <math>13</math>, and [[point]]s <math>E</math> and <math>F</math> are exterior to the square such that <math>BE=DF=5</math> and <math>AE=CF=12</math>. Find <math>EF^{2}</math>.
 +
<asy>unitsize(0.2 cm);
  
<div style="text-align:center;">[[Image:2007 AIME II-3.png]]</div>
+
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>
 +
 
 +
__TOC__
  
 
== Solution ==
 
== Solution ==
  
 +
===Solution 1===
  
== '''Solution 1.''' ==
+
Let <math>\angle FCD = \alpha</math>, so that <math>FB = \sqrt{12^2 + 13^2 + 2\cdot12\cdot13\sin(\alpha)} = \sqrt{433}</math>. By the diagonal, <math>DB = 13\sqrt{2}, DB^2 = 338</math>.
 +
 
 +
<b>The sum of the squares of the sides of a parallelogram is the sum of the squares of the diagonals.</b>
 +
<cmath>EF^2 = 2\cdot(5^2 + 433) - 338 = 578.</cmath>
 +
 
 +
=== Solution 2 ===
  
 
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 [[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>.
  
<div style='text-align:center;'>[[Image:2007 AIME II-3b.PNG]]</div>
+
<asy>unitsize(0.25 cm);
 +
 
 +
pair A, B, C, D, E, F, G, H;
  
<math>\overline{EF}</math> is the diagonal of the square, with length <math>17\sqrt{2}</math>; the answer is <math>EF^2 = (17\sqrt{2})^2 = 578</math>.
+
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);
  
== '''Solution 2.''' ==
+
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>
  
 +
<math>\overline{EF}</math> is the diagonal of the square, with length <math>17\sqrt{2}</math>; the answer is <math>EF^2 = (17\sqrt{2})^2 = 578</math>.
  
 +
=== Solution 3 ===
 
A slightly more analytic/brute-force approach:
 
A slightly more analytic/brute-force approach:
  
Line 25: Line 72:
 
Drop perpendiculars from <math>E</math> and <math>F</math> to <math>I</math> and <math>J</math>, respectively; construct right triangle <math>EKF</math> with right angle at K and <math>EK || BC</math>. Since <math>2[CDF]=DF*CF=CD*JF</math>, we have <math>JF=5\times12/13 = \frac{60}{13}</math>. Similarly, <math>EI=\frac{60}{13}</math>. Since <math>\triangle DJF \sim \triangle DFC</math>, we have <math>DJ=\frac{5JF}{12}=\frac{25}{13}</math>.
 
Drop perpendiculars from <math>E</math> and <math>F</math> to <math>I</math> and <math>J</math>, respectively; construct right triangle <math>EKF</math> with right angle at K and <math>EK || BC</math>. Since <math>2[CDF]=DF*CF=CD*JF</math>, we have <math>JF=5\times12/13 = \frac{60}{13}</math>. Similarly, <math>EI=\frac{60}{13}</math>. Since <math>\triangle DJF \sim \triangle DFC</math>, we have <math>DJ=\frac{5JF}{12}=\frac{25}{13}</math>.
  
Now, we see that <math>FK=DC-(DJ+IB)=DC-2DJ=13-\frac{50}{13}=\frac{119}{13}</math>. Also, <math>EK=BC+(JF+IE)=BC+2JF=13+\frac{120}{13}=\frac{289}{13}</math>. By the Pythagorean Theorem, we have <math>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}</math>. Therefore, <math>EF^2=(17\sqrt{2})^2=578</math>.
+
Now, we see that <math>FK=DC-(DJ+IB)=DC-2DJ=13-\frac{50}{13}=\frac{119}{13}</math>. Also, <math>EK=BC+(JF+IE)=BC+2JF=13+\frac{120}{13}=\frac{289}{13}</math>. By the Pythagorean Theorem, we have <math>EF=\sqrt{\left(\frac{289}{13}\right)^2+\left(\frac{119}{13} \right)^2}=\frac{\sqrt{(17^2)(17^2+7^2)}}{13}</math><math>=\frac{17\sqrt{338}}{13}=\frac{17(13\sqrt{2})}{13}=17\sqrt{2}</math>. Therefore, <math>EF^2=(17\sqrt{2})^2=578</math>.
 +
 
 +
===Solution 4===
 +
Based on the symmetry, we know that <math>F</math> is a reflection of <math>E</math> across the center of the square, which we will denote as <math>O</math>. Since <math>\angle BEA</math> and <math>\angle AOB</math> are right, we can conclude that figure <math>AOBE</math> is a cyclic quadrilateral. Pythagorean Theorem yields that <math>BO=AO=\frac{13\sqrt{2}}{2}</math>. Now, using Ptolemy's Theorem, we get that
 +
<cmath>AO\cdot BE + BO\cdot AE = AB\cdot AO</cmath>
 +
<cmath>\frac{13\sqrt{2}}{2}\cdot 5+\frac{13\sqrt{2}}{2}\cdot 12 = 13\cdot OE</cmath>
 +
<cmath>OE=\frac{17\sqrt{2}}{2}</cmath>
 +
Now, since we stated in the first step that <math>F</math> is a reflection of <math>E</math> across <math>O</math>, we can say that <math>EF=2EO=17\sqrt{2}</math>. This gives that <cmath>EF^2=(17\sqrt{2})^2=578</cmath> AWD with this bash solution
 +
 
 +
===Solution 5 (Ptolemy's Theorem)===
 +
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===
 +
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 <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>
 +
 
 +
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>
 +
 
 +
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 ==
Line 31: Line 102:
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 20:46, 3 August 2020

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS