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1984 AHSME Problems/Problem 27

Revision as of 18:44, 10 December 2024 by Elpianista227 (talk | contribs) (→‎Solution 2 (single-variable))

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 $x=\sqrt[3]{2}, \boxed{\text{C}}$.

Solution 2 (single-variable)

Let $AC = x$ and draw $FE||BD$. Since $FE$ is the median of a right triangle, it follows that $FE = AE = CE = \frac{x}{2}$. Then, since $\Delta EFC$ and $\Delta DBC$ are isoceles triangles, then they are considered similar. Therefore, $\frac{1}{2x} = BC$ through similarity ratios. Finally, using the Pythagorean Theorem and the fact that $AD = x - 1$ gives $AB^2 = \frac{4 - x^4}{x^2}$ and also 

   \[(x - 1)^2 + \frac{4 - x^4}{x^2} = 1\]
   \[x^2 - 2x + 1 + \frac{4 - x^4}{x^2} = 1\]
   \[x^2 - 2x + \frac{4 - x^4}{x^2} = 0\]
   \[x^4 - 2x^3 + 4 - x^4 = 0\]
   \[-2x^3 + 4 = 0\]
   \[-2(x^3 - 2) = 0; x^3 = 2\]
   Therefore, $AC^3 = 2, \boxed{\text{C}}$.

~elpianista227

See Also

1984 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 26 		Followed by
Problem 28
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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