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Difference between revisions of "2002 AMC 10A Problems/Problem 23"

(→‎Simpler Solution)
(Asymptope addition)
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==Solution==
 
==Solution==
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First, we draw an altitude to <math>BC</math> from <math>E</math>. Let it intersect at <math>M</math>. As <math>\triangle BEC</math> is isosceles, we immediately get <math>MB=MC=6</math>, so the altitude is <math>8</math>. Now, let <math>AB=CD=x</math>. Using the Pythagorean Theorem on <math>\triangle EMA</math>, we find <math>AE=\sqrt{x^2+12x+100}</math>. From symmetry, <math>DE=\sqrt{x^2+12x+100}</math> as well. Now, we use the fact that the perimeter of <math>\triangle AED</math> is twice the perimeter of <math>\triangle BEC</math>.  
+
First, we draw an altitude to <math>BC</math> from <math>E</math>. Let it intersect at <math>M</math>. As <math>\triangle BEC</math> is isosceles, we immediately get <math>MB=MC=6</math>, so the altitude is <math>8</math>. Now, let <math>AB=CD=x</math>. Using the Pythagorean Theorem on <math>\triangle EMA</math>, we find <math>AE=\sqrt{x^2+12x+100}</math>. From symmetry, <math>DE=\sqrt{x^2+12x+100}</math> as well. Now, we use the fact that the perimeter of <math>\triangle AED</math> is twice the perimeter of <math>\triangle BEC</math>.
 +
 
 +
<asy>
 +
unitsize(0.25 cm);
 +
 
 +
pair A, B, C, D, E, M;
 +
 
 +
A = (0,0);
 +
B = (9,0);
 +
C = (21,0);
 +
D = (30,0);
 +
E = (15,-8);
 +
M = (15,0);
 +
 
 +
draw(A--D--E--cycle);
 +
draw(B--E);
 +
draw(M--E);
 +
draw(C--E);
 +
 
 +
label("$A$", A, N);
 +
label("$B$", B, N);
 +
label("$C$", C, N);
 +
label("$D$", D, N);
 +
label("$E$", E, S);
 +
label("$M$", M, N);
 +
</asy>
  
 
We have <math>2\sqrt{x^2+12x+100}+2x+12=2(32)</math> so <math>\sqrt{x^2+12x+100}=26-x</math>. Squaring both sides, we have <math>x^2+12x+100=676-52x+x^2</math> which nicely rearranges into <math>64x=576\rightarrow{x=9}</math>. Hence, AB is 9 so our answer is <math>\boxed{\text{(D)}}</math>.
 
We have <math>2\sqrt{x^2+12x+100}+2x+12=2(32)</math> so <math>\sqrt{x^2+12x+100}=26-x</math>. Squaring both sides, we have <math>x^2+12x+100=676-52x+x^2</math> which nicely rearranges into <math>64x=576\rightarrow{x=9}</math>. Hence, AB is 9 so our answer is <math>\boxed{\text{(D)}}</math>.

Revision as of 18:24, 8 August 2022

Problem 23

Points $A,B,C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$. Point $E$ is not on the line, and $BE = CE = 10$. The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$. Find $AB$.

$\text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 19/2$

Solution

First, we draw an altitude to $BC$ from $E$. Let it intersect at $M$. As $\triangle BEC$ is isosceles, we immediately get $MB=MC=6$, so the altitude is $8$. Now, let $AB=CD=x$. Using the Pythagorean Theorem on $\triangle EMA$, we find $AE=\sqrt{x^2+12x+100}$. From symmetry, $DE=\sqrt{x^2+12x+100}$ as well. Now, we use the fact that the perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$.

[asy] unitsize(0.25 cm); pair A, B, C, D, E, M; A = (0,0); B = (9,0); C = (21,0); D = (30,0); E = (15,-8); M = (15,0); draw(A--D--E--cycle); draw(B--E); draw(M--E); draw(C--E); label("$A$", A, N); label("$B$", B, N); label("$C$", C, N); label("$D$", D, N); label("$E$", E, S); label("$M$", M, N); [/asy]

We have $2\sqrt{x^2+12x+100}+2x+12=2(32)$ so $\sqrt{x^2+12x+100}=26-x$. Squaring both sides, we have $x^2+12x+100=676-52x+x^2$ which nicely rearranges into $64x=576\rightarrow{x=9}$. Hence, AB is 9 so our answer is $\boxed{\text{(D)}}$.

Simpler Solution 2

Let $M$ be the foot of the altitude from $E$ to $BC.$ Then $MB=MC=6$ because $\triangle BEC$ is isosceles. By the Pythagorean triple $(6,8,10)$ the altitude is $8.$ Since $(8,15,17)$ is the only primitive Pythagorean triple with leg $8,$ we test $AE=DE=17,AM=DM=15.$ Since $2(10+10+12)=(17+17+2\cdot 15)$ this works, giving us $AB=15-6=\boxed{\text{(D)}\ 9}.$

~dolphin7

See Also

2002 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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