Difference between revisions of "2002 AMC 10A Problems/Problem 23"
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==Solution== | ==Solution== | ||
− | 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>. | ||
− | ==Simpler Solution== | + | ==Simpler Solution 2== |
− | Let <math>M</math> be the foot of the altitude from <math>E</math> to <math>BC.</math> Then <math>MB=MC=6</math> because <math>\triangle BEC</math> is isosceles. By the Pythagorean triple <math>(6,8,10)</math> the altitude is <math>8.</math> Since <math>(8,15,17)</math> is the only primitive Pythagorean triple with leg <math>8,</math> we test <math>AE=DE=17,AM=DM=15.</math> Since <math>2(10+10+12)=(17+17+2\cdot 15)</math> this works, giving us <math>AB=15- | + | Let <math>M</math> be the foot of the altitude from <math>E</math> to <math>BC.</math> Then <math>MB=MC=6</math> because <math>\triangle BEC</math> is isosceles. By the Pythagorean triple <math>(6,8,10)</math> the altitude is <math>8.</math> Since <math>(8,15,17)</math> is the only primitive Pythagorean triple with leg <math>8,</math> we test <math>AE=DE=17,AM=DM=15.</math> Since <math>2(10+10+12)=(17+17+2\cdot 15)</math> this works, giving us <math>AB=15-6=\boxed{\text{(D)}\ 9}.</math> |
~dolphin7 | ~dolphin7 | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://www.youtube.com/watch?v=HJkfO6vuIwg ~David | ||
==See Also== | ==See Also== |
Latest revision as of 20:38, 19 July 2023
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 from . Let it intersect at . As is isosceles, we immediately get , so the altitude is . Now, let . Using the Pythagorean Theorem on , 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 .
Simpler Solution 2
Let be the foot of the altitude from to Then because is isosceles. By the Pythagorean triple the altitude is Since is the only primitive Pythagorean triple with leg we test Since this works, giving us
~dolphin7
Video Solution
https://www.youtube.com/watch?v=HJkfO6vuIwg ~David
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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