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1986 AIME Problems/Problem 9

Revision as of 21:30, 8 April 2008 by Dgreenb801 (talk | contribs) (→‎Solution 1)

Problem

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

Solution

Solution 1

1986 AIME-9.png

Let the points at which the segments hit the triangle be called $D, D', E, E', F, F'$ as shown above. All three smaller triangles and the larger triangle are similar ($\triangle ABC \sim \triangle DPD' \sim \triangle PEE' \sim \triangle F'PF$). This is easy to find using repeated alternate interior angles. The remaining three sections are parallelograms, which is also simple to see by the parallel lines.

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:

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

Doing the same with $\triangle PEE'$, we find that $PE' =510 - \frac{17}{15}d$. Now, $d = PD + PE' = 510 - \frac{6}{5}d + 510 - \frac{17}{15}d \Longrightarrow d\left(\frac{50}{15}\right) = 1020 \Longrightarrow d = 306$.

Solution 2

Define the points the same as above.

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

Key theorem: 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:

$\frac {c + e + d}{A} = \left(\frac {x}{BC}\right)^2$, $\displaystyle \frac {b + c + d}{A} \displaystyle = \left(\frac {x}{AC}\right)^2$, $\frac {a + b + f}{A} = \left(\frac {x}{AB}\right)^2$ 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 $\displaystyle \frac {f}{A} = \left(\frac {AC - x}{AC}\right)^2 = \left(1 - \frac {x}{AC}\right)^2$, by symmetry, we have $\displaystyle \frac {d}{A} = \left(1 - \frac {x}{AB}\right)^2$ and $\frac {b}{A} = \left(1 - \frac {x}{BC}\right)^2$

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

See also

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