# 1984 AHSME Problems/Problem 27

## Problem

In $\triangle ABC$, $D$ is on $AC$ and $F$ is on $BC$. Also, $AB\perp AC$, $AF\perp BC$, and $BD=DC=FC=1$. Find $AC$.

$\mathrm{(A) \ }\sqrt{2} \qquad \mathrm{(B) \ }\sqrt{3} \qquad \mathrm{(C) \ } \sqrt[3]{2} \qquad \mathrm{(D) \ }\sqrt[3]{3} \qquad \mathrm{(E) \ } \sqrt[4]{3}$

## Solution

$[asy] unitsize(4cm); draw((0,0)--(2^(1/3),0)); draw((0,0)--(0,16^(1/3)-4^(1/3))); draw((2^(1/3),0)--(0,16^(1/3)-4^(1/3))); draw((0,16^(1/3)-4^(1/3))--(2^(1/3)-1,0)); draw((0,0)--(0.445861,0.602489)); label("A",(0,0),W); label("B",(0,16^(1/3)-4^(1/3)),N); label("C",(2^(1/3),0),E); label("D",(2^(1/3)-1,0),SE); label("F",(0.445861,0.602489),NE); label("1",((2^(1/3)+.445861)/2,.602489/2),NE); label("1",(2^(1/3)-.5,0),S); label("1", (.13, .932441/2), NE); label("x-1",(.13,0),S); label("\sqrt{2x-x^2}",(0,.932441/2),W); label("y",(.445861/2,(.932441+.602489)/2),NE); label("\sqrt{y}",(.445861/2,.602489/2),E); [/asy]$

Let $AC=x$ and $BF=y$. We have $AFC\sim BFA$ by AA, so $\frac{AF}{FC}=\frac{BF}{AF}$. Substituting in known values gives $\frac{AF}{1}=\frac{y}{AF}$, so $AF=\sqrt{y}$. Also, $AD=x-1$, and using the Pythagorean Theorem on $\triangle ABD$, we have $AB^2+(x-1)^2=1^2$, so $AB=\sqrt{2x-x^2}$. Using the Pythagorean Theorem on $\triangle AFC$ gives $y+1=x^2$, or $y=x^2-1$. Now, we use the Pythagorean Theorem on $\triangle AFB$ to get $y^2+y=2x-x^2$. Substituting $y=x^2-1$ into this gives $(x^2-1)^2+x^2-1=2x-x^2$, or $x^4-2x^2+1+x^2-1=2x-x^2$. Simplifying this and moving all of the terms to one side gives $x^4-2x=0$, and since $x\not=0$, we can divide by $x$ to get $x^3-2=0$, from which we find that $x=\sqrt[3]{2}, \boxed{\text{C}}$.