2002 AMC 10A Problems/Problem 23: Difference between revisions

Revision as of 10:13, 4 July 2013

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

Revision as of 01:39, 24 February 2010 view source Xrayxd (talk \| contribs) 5 edits m →Solution ← Older edit		Revision as of 10:13, 4 July 2013 view source Nathan wailes (talk \| contribs) 2,960 edits No edit summary Newer edit →
Line 13:		Line 13:

	[[Category:Introductory Geometry Problems]]		[[Category:Introductory Geometry Problems]]
			{{MAA Notice}}

Page

Toolbox

Search

2002 AMC 10A Problems/Problem 23: Difference between revisions

Revision as of 10:13, 4 July 2013

Problem 23

Solution

See Also