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...) |
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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 | + | 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 02:39, 24 February 2010
Problem 23
Points and
lie on a line, in that order, with
and
. Point
is not on the line, and
. The perimeter of
is twice the perimeter of
. Find
.
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 . Using the Pythagorean Theorem on triangle EMA, we find
. From symmetry,
as well. Now, we use the fact that the perimeter of
is twice the perimeter of
.
We have so
. Squaring both sides, we have
which nicely rearranges into
. Hence, AB is 9 so our answer is
.
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 |