# Difference between revisions of "1988 AIME Problems/Problem 12"

## Problem

Let $P$ be an interior point of triangle $ABC$ and extend lines from the vertices through $P$ to the opposite sides. Let $a$, $b$, $c$, and $d$ denote the lengths of the segments indicated in the figure. Find the product $abc$ if $a + b + c = 43$ and $d = 3$.

## Solution 1

Call the cevians AD, BE, and CF. Using area ratios ( $\triangle PBC$ and $\triangle ABC$ have the same base), we have: $\frac {d}{a + d} = \frac {[PBC]}{[ABC]}$

Similarily, $\frac {d}{b + d} = \frac {[PCA]}{[ABC]}$ and $\frac {d}{c + d} = \frac {[PAB]}{[ABC]}$.

Then, $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = \frac {[PBC]}{[ABC]} + \frac {[PCA]}{[ABC]} + \frac {[PAB]}{[ABC]} = \frac {[ABC]}{[ABC]} = 1$

The identity $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d} = 1$ is a form of Ceva's Theorem.

Plugging in $d = 3$, we get $$\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1$$ $$3[(a + 3)(b + 3) + (b + 3)(c + 3) + (c + 3)(a + 3)] = (a+3)(b+3)(c+3)$$ $$3(ab + bc + ca) + 18(a + b + c) + 81 = abc + 3(ab + bc + ca) + 9(a + b + c) + 27$$ $$9(a + b + c) + 54 = abc=\boxed{441}$$

## Solution 2

Let $A,B,C$ be the weights of the respective vertices. We see that the weights of the feet of the cevians are $A+B,B+C,C+A$. By mass points, we have that: $$\dfrac{a}{3}=\dfrac{B+C}{A}$$ $$\dfrac{b}{3}=\dfrac{C+A}{B}$$ $$\dfrac{c}{3}=\dfrac{A+B}{C}$$

If we add the equations together, we get $\frac{a+b+c}{3}=\frac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B}{ABC}=\frac{43}{3}$

If we multiply them together, we get $\frac{abc}{27}=\frac{A^2B+A^2C+B^2A+B^2C+C^2A+C^2B+2ABC}{ABC}=\frac{49}{3} \implies abc=\boxed{441}$

## Solution 3

You can use mass points to derive $\frac {d}{a + d} + \frac {d}{b + d} + \frac {d}{c + d}=1.$ Plugging it in yields $\frac{3}{a + 3} + \frac{3}{b + 3} + \frac{3}{c+3} = 1.$ We proceed as we did in Solution 1 - however, to make the equation look less messy, we do the substitution $a'=a+3,b'=b+3,c'=c+3.$

Then we have $\frac{3}{a'}+\frac{3}{b'}+\frac{3}{c'}=1.$ Clearing fractions gives us $a'b'c'=3a'b'+3b'c'+3c'a'\to a'b'c'-3a'b'-3b'c'-3c'a'=0.$ Factoring yields $(a'-3)(b'-3)(c'-3)=9(a'+b'+c')-27,$ and the left hand side looks suspiciously like what we want to find. (It is.)

Substituting yields our answer as $9\cdot 52-27=\boxed{441}.$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 