Difference between revisions of "2002 AMC 10A Problems/Problem 23"
(→Simpler Solution 2) |
|||
(7 intermediate revisions by 5 users not shown) | |||
Line 5: | Line 5: | ||
==Solution== | ==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 <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 <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 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=15-6=\boxed{\text{(D)}\ 9}.</math> | ||
+ | |||
+ | ~dolphin7 | ||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://www.youtube.com/watch?v=HJkfO6vuIwg ~David | ||
==See Also== | ==See Also== |
Latest revision as of 21:38, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.