Difference between revisions of "2015 AIME II Problems/Problem 7"
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==Solution 2== | ==Solution 2== | ||
− | + | <asy> | |
− | pair A,B,C,E, | + | 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> | </asy> | ||
Similar triangles can also solve the problem. | Similar triangles can also solve the problem. |
Revision as of 09:25, 30 March 2015
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 1
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
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 .
- solution by abvenkgoo
Solution 3
Heron's Formula gives so the altitude from to has length
Now, draw a parallel to from , intersecting at . Then in parallelogram , and so . Clearly, and are similar triangles, and so their altitudes have lengths proportional to their corresponding base sides, and so Solving gives , so the answer is .
- solution by suli
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.