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

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(→‎Solution 1)
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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 [[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;
 +
 
 +
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>
  
 
<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>.
 
<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 2 ===
 
=== Solution 2 ===

Revision as of 14:49, 1 May 2015

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

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 2

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, $EF^2=(17\sqrt{2})^2=578$.

See also

2007 AIME II (Problems • Answer Key • Resources)
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

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