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1986 AIME Problems/Problem 9

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Problem

In $\triangle ABC$ , $AB= 425$ , $BC=450$ , and $AC=510$ . An interior point $P$ is then drawn, and segments are drawn through $P$ parallel to the sides of the triangle. If these three segments are of an equal length $d$ , find $d$ .

Contents

1 Problem
2 Solution
3 See also

Solution

Solution 1

size(200);pathpen = black; pointpen = black +linewidth(0.6); pen s = fontsize(10);pair C=(0,0),A=(510,0),B=IP(circle(C,450),c...

Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. As a result of the lines being parallel, all three smaller triangles and the larger triangle are similar ( $\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$ ). The remaining three sections are parallelograms.

Since $PDAF'$ is a parallelogram, we find $PF' = AD$ , and similarly $PE = BD'$ . So $d = PF' + PE = AD + BD' = 425 - DD'$ . Thus $DD' = 425 - d$ . By the same logic, $EE' = 450 - d$ .

Since $\triangle DPD' \sim \triangle ABC$ , we have the proportion:

$\frac{425-d}{425} = \frac{PD}{510} \Longrightarrow PD = 510 - \frac{510}{425}d = 510 - \frac{6}{5}d$

Doing the same with $\triangle PEE'$ , we find that $PE' =510 - \frac{17}{15}d$ . Now, $d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d...$ .

Solution 2

Define the points the same as above.

Let $[CE'PF] = a$ , $[E'EP] = b$ , $[BEPD'] = c$ , $[D'PD] = d$ , $[DAF'P] = e$ and $[F'D'P] = f$

The key theorem we apply here is that the ratio of the areas of 2 similar triangles is the ratio of a pair of corresponding sides squared.

Let the length of the segment be $x$ and the area of the triangle be $A$ , using the theorem, we get:

$\frac {c + e + d}{A} = \left(\frac {x}{BC}\right)^2$ , $\frac {b + c + d}{A}= \left(\frac {x}{AC}\right)^2$ , $\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2$ adding all these together and using $a + b + c + d + e + f = A$ we get $\frac {f + d + b}{A} + 1 = x^2*\left(\frac {1}{BC^2} + \frac {1}{AC^2} + \frac {1}{AB^2}\right)$

Using corresponding angles from parallel lines, it is easy to show that $\triangle ABC \sim \triangle F'PF$ , since $ADPF'$ and $CFPE'$ are parallelograms, it is easy to show that $FF' = AC - x$

Now we have the side length ratio, so we have the area ratio $\frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2$ , by symmetry, we have $\frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2$ and $\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2$

Substituting these into our initial equation, we have $1 + \sum_{cyc}\left(1 - \frac {x}{AB}\right) - \frac {x^2}{AB^2} = 0$ $1 + \sum_{cyc}1 - 2*\frac {x}{AB} = 0$ $\frac {2}{\frac {1}{AB} + \frac {1}{BC} + \frac {1}{CA}} = x$ answer follows after some hideous computation.

Solution 3

Refer to the diagram; let $a^2=[E'EP]$ , $b^2=[D'DP]$ , and $c^2=[F'FP]$ . Now, note that $[E'BD]$ , $[D'DP]$ , and $[E'EP]$ are similar, so through some similarities we find that $\frac{E'P}{PD}=\frac{a}{b}\implies\frac{E'D}{PD}=\frac{a+b}{b}\implies[E'BD]=b^2\left(\frac{a+b}{b}\right)^2=(a+b)^2$ . Similarly, we find that $[D'AF]=(b+c)^2$ and $[F'CE]=(c+a)^2$ , so $[ABC]=(a+b+c)^2$ . Now, again from similarity, it follows that $\frac{d}{510}=\frac{a+b}{a+b+c}$ , $\frac{d}{450}=\frac{b+c}{a+b+c}$ , and $\frac{d}{425}=\frac{c+a}{a+b+c}$ , so adding these together, simplifying, and solving gives $d=\frac{2}{\frac{1}{425}+\frac{1}{450}+\frac{1}{510}}=\frac{10}{\frac{1}{85}+\frac{1}{90}+\frac{1}{102}}=\frac{10}{\frac{1}{5...$ $=\frac{10}{\frac{10}{306}}=\boxed{306}$ .

See also

1986 AIME (Problems • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1986_AIME_Problems/Problem_9"

Category: Intermediate Geometry Problems

Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.

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