Difference between revisions of "2008 AIME I Problems/Problem 2"
(→Solution) |
|||
(One intermediate revision by the same user not shown) | |||
Line 22: | Line 22: | ||
Let the height of <math>GXY</math> be <math>h</math>. By the similarity, <math>\dfrac{h}{6} = \dfrac{h + 10}{10}</math>, we get <math>h = 15</math>. Thus, the height of <math>GEM</math> is <math>h + 10 = \boxed{025}</math>. | Let the height of <math>GXY</math> be <math>h</math>. By the similarity, <math>\dfrac{h}{6} = \dfrac{h + 10}{10}</math>, we get <math>h = 15</math>. Thus, the height of <math>GEM</math> is <math>h + 10 = \boxed{025}</math>. | ||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
== See also == | == See also == |
Latest revision as of 19:37, 29 December 2020
Problem
Square has sides of length
units. Isosceles triangle
has base
, and the area common to triangle
and square
is
square units. Find the length of the altitude to
in
.
Solution 1
Note that if the altitude of the triangle is at most , then the maximum area of the intersection of the triangle and the square is
.
This implies that vertex G must be located outside of square
.
![[asy] pair E=(0,0), M=(10,0), I=(10,10), A=(0,10); draw(A--I--M--E--cycle); pair G=(5,25); draw(G--E--M--cycle); label("\(G\)",G,N); label("\(A\)",A,NW); label("\(I\)",I,NE); label("\(M\)",M,NE); label("\(E\)",E,NW); label("\(10\)",(M+E)/2,S); [/asy]](http://latex.artofproblemsolving.com/6/2/1/621cda5459b0a04c3bb47fbec28d266740d0a08b.png)
Let meet
at
and let
meet
at
. Clearly,
since the area of trapezoid
is
. Also,
.
Let the height of be
. By the similarity,
, we get
. Thus, the height of
is
.
See also
2008 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
this could have been a #1-#5 on the amc 10 lol