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

(New page: == Problem 23 == Points <math>A,B,C</math> and <math>D</math> lie on a line, in that order, with <math>AB = CD</math> and <math>BC = 12</math>. Point <math>E</math> is not on the line, and...)
 
m (→‎Solution)
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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 <math>AB=CD=x</math>. Using the Pythagorean Theorem on triangle EMA, 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 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 <math>AB=CD=x</math>. Using the Pythagorean Theorem on triangle EMA, 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>.  
  
−
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 nice 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>.
  
 
==See Also==
 
==See Also==

Revision as of 01:39, 24 February 2010

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 $\triangle BEC$.

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 $\boxed{\text{(D)}}$.

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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