Difference between revisions of "2015 AIME II Problems/Problem 7"
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==Solution #2== | ==Solution #2== | ||
Diagram: | Diagram: | ||
− | + | <asy> | |
− | ( | + | unitsize(35); |
+ | draw(Circle((-1,0),1)); | ||
+ | draw(Circle((4,0),4)); | ||
+ | pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; | ||
+ | A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); | ||
+ | label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4)); | ||
+ | draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); | ||
+ | dot(O_1);dot(O_2); | ||
+ | draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); | ||
+ | </asy> | ||
Similar triangles can also solve the problem. | Similar triangles can also solve the problem. |
Revision as of 12:42, 29 March 2015
Contents
[hide]Problem
Triangle has side lengths
,
, and
. Rectangle
has vertex
on
, vertex
on
, and vertices
and
on
. In terms of the side length
, the area of
can be expressed as the quadratic polynomial
Area() =
.
Then the coefficient , where
and
are relatively prime positive integers. Find
.
Solution
If , the area of rectangle
is
, so
and . If
, we can reflect
over PQ,
over
, and
over
to completely cover rectangle
, so the area of
is half the area of the triangle. Using Heron's formula, since
,
so
and
so the answer is .
Solution #2
Diagram:
Similar triangles can also solve the problem.
First, solve for the area of the triangle. . This can be done by Heron's Formula or placing an
right triangle on
and solving. (The
side would be collinear with line
)
After finding the area, solve for the altitude to . Let
be the intersection of the altitude from
and side
. Then
.
Solving for
using the Pythagorean Formula, we get
. We then know that
.
Now consider the rectangle . Since
is collinear with
and parallel to
,
is parallel to
meaning
is similar to
.
Let be the intersection between
and
. By the similar triangles, we know that
. Since
. We can solve for
and
in terms of
. We get that
and
.
Let's work with . We know that
is parallel to
so
is similar to
. We can set up the proportion:
. Solving for
,
.
We can solve for then since we know that
and
.
Therefore, .
This means that .
See also
2015 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.