Difference between revisions of "2015 AIME II Problems/Problem 7"

Revision as of 12:55, 29 March 2015

Problem

Triangle $ABC$ has side lengths $AB = 12$ , $BC = 25$ , and $CA = 17$ . Rectangle $PQRS$ has vertex $P$ on $\overline{AB}$ , vertex $Q$ on $\overline{AC}$ , and vertices $R$ and $S$ on $\overline{BC}$ . In terms of the side length $PQ = w$ , the area of $PQRS$ can be expressed as the quadratic polynomial

Area( $PQRS$ ) = $\alpha w - \beta \cdot w^2$ .

Then the coefficient $\beta = \frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

If $\omega = 25$ , the area of rectangle $PQRS$ is $0$ , so

$\alpha\omega - \beta\omega^2 = 25\alpha - 625\beta = 0$

and $\alpha = 25\beta$ . If $\omega = \frac{25}{2}$ , we can reflect $APQ$ over PQ, $PBS$ over $PS$ , and $QCR$ over $QR$ to completely cover rectangle $PQRS$ , so the area of $PQRS$ is half the area of the triangle. Using Heron's formula, since $s = \frac{12 + 17 + 25}{2} = 27$ ,

$[ABC] = \sqrt{27 \cdot 15 \cdot 10 \cdot 2} = 90$

so

$45 = \alpha\omega - \beta\omega^2 = \frac{625}{2} \beta - \beta\frac{625}{4} = \beta\frac{625}{4}$

and

$\beta = \frac{180}{625} = \frac{36}{125}$

so the answer is $m + n = 36 + 125 = \boxed{161}$ .

Solution #2

Similar triangles can also solve the problem.

First, solve for the area of the triangle. $[ABC] = 90$ . This can be done by Heron's Formula or placing an $8-15-17$ right triangle on $BC$ and solving. (The $8$ side would be collinear with line $AB$ )

After finding the area, solve for the altitude to $BC$ . Let $E$ be the intersection of the altitude from $A$ and side $BC$ . Then $AE = \frac{36}{5}$ . Solving for $BE$ using the Pythagorean Formula, we get $BE = \frac{48}{5}$ . We then know that $CE = \frac{77}{5}$ .

Now consider the rectangle $PQRS$ . Since $SR$ is collinear with $BC$ and parallel to $PQ$ , $PQ$ is parallel to $BC$ meaning $\Delta APQ$ is similar to $\Delta ABC$ .

Let $F$ be the intersection between $AE$ and $PQ$ . By the similar triangles, we know that $\frac{PF}{FQ}=\frac{BE}{EC} = \frac{48}{77}$ . Since $PF+FQ=PQ=\omega$ . We can solve for $PF$ and $FQ$ in terms of $\omega$ . We get that $PF=\frac{48}{125} \omega$ and $FQ=\frac{77}{125} \omega$ .

Let's work with $PF$ . We know that $PQ$ is parallel to $BC$ so $\Delta APF$ is similar to $\Delta ABE$ . We can set up the proportion:

$\frac{AF}{PF}=\frac{AE}{BE}=\frac{3}{4}$ . Solving for $AF$ , $AF = \frac{3}{4} PF = \frac{3}{4} \cdot \frac{48}{125} \omega = \frac{36}{125} \omega$ .

We can solve for $PS$ then since we know that $PS=FE$ and $FE= AE - AF = \frac{36}{5} - \frac{36}{125} \omega$ .

Therefore, $[PQRS] = PQ \cdot PS = \omega (\frac{36}{5} - \frac{36}{125} \omega) = \frac{36}{5}\omega - \frac{36}{125} \omega^2$ .

This means that $\beta = \frac{36}{125} \Rightarrow (m,n) = (36,125) \Rightarrow m+n = \boxed{161}$ .

- solution by abvenkgoo

@@ Line 28: / Line 28: @@
 ==Solution #2==
-Diagram:
-<asy>
-unitsize(35);
-pair A,B,C,E,F,P,Q,R,S;
-B=(0,0); S=(24/5,0); E = (48/5,0); R=(173/10,0);C=(25,0);A=(48/5,36/5);F=(48/5,18/5);P=(24/5,18/5);Q=(173/10,18/5);
-</asy>
 Similar triangles can also solve the problem.

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Difference between revisions of "2015 AIME II Problems/Problem 7"

Revision as of 12:55, 29 March 2015

Contents

Problem

Solution

Solution #2

See also