# 1985 AHSME Problems/Problem 28

## Problem

In $\triangle ABC$, we have $\angle C=3\angle A, a=27$ and $c=48$. What is $b$? $[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("A", A, SW); label("B", B, SE); label("C", C, N); label("a", B--C, dir(B--C)*dir(-90)); label("b", A--C, dir(C--A)*dir(-90)); label("c", A--B, dir(A--B)*dir(-90));[/asy]$ $\mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \ } 37 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ }\text{not uniquely determined}$

## Solution 1

From the Law of Sines, we have $\frac{\sin(A)}{a}=\frac{\sin(C)}{c}$, or $\frac{\sin(A)}{27}=\frac{\sin(3A)}{48}\implies 9\sin(3A)=16\sin(A)$.

We now need to find an identity relating $\sin(3A)$ and $\sin(A)$. We have $\sin(3A)=\sin(2A+A)=\sin(2A)\cos(A)+\cos(2A)\sin(A)$ $=2\sin(A)\cos^2(A)+\sin(A)\cos^2(A)-\sin^3(A)=3\sin(A)\cos^2(A)-\sin^3(A)$ $=3\sin(A)(1-\sin^2(A))-\sin^3(A)=3\sin(A)-4\sin^3(A)$.

Thus we have $9(3\sin(A)-4\sin^3(A))=27\sin(A)-36\sin^3(A)=16\sin(A)$ $\implies 36\sin^3(A)=11\sin(A)$.

Therefore, $\sin(A)=0, \frac{\sqrt{11}}{6},$ or $-\frac{\sqrt{11}}{6}$. Notice that we must have $0^\circ because otherwise $A+3A>180^\circ$. We can therefore disregard $\sin(A)=0$ because then $A=0$ and also we can disregard $\sin(A)=-\frac{\sqrt{11}}{6}$ because then $A$ would be in the third or fourth quadrants, much greater than the desired range.

Therefore, $\sin(A)=\frac{\sqrt{11}}{6}$, and $\cos(A)=\sqrt{1-\left(\frac{\sqrt{11}}{6}\right)^2}=\frac{5}{6}$. Going back to the Law of Sines, we have $\frac{\sin(A)}{27}=\frac{\sin(B)}{b}=\frac{\sin(\pi-3A-A)}{b}=\frac{\sin(4A)}{b}$.

We now need to find $\sin(4A)$. $\sin(4A)=\sin(2\cdot2A)=2\sin(2A)\cos(2A)$ $=2(2\sin(A)\cos(A))(\cos^2(A)-\sin^2(A))$ $=2\cdot2\cdot\frac{\sqrt{11}}{6}\cdot\frac{5}{6}\left(\left(\frac{5}{6}\right)^2-\left(\frac{\sqrt{11}}{6}\right)^2\right)=\frac{35\sqrt{11}}{162}$.

Therefore, $\frac{\frac{\sqrt{11}}{6}}{27}=\frac{\frac{35\sqrt{11}}{162}}{b}\implies b=\frac{6\cdot27\cdot35}{162}=35, \boxed{\text{B}}$.

## Solution 2

Let angle $A$ be equal to $x$ degrees. Then angle $C$ is equal to $3x$ degrees, and angle $B$ is equal to $180-4x$ degrees. Let $D$ be a point on side $AB$ such that $\angle ACD$ is equal to $x$ degrees. Because $2x+180-4x+\angle CDB=180$, angle $CDB$ is equal to $2x$ degrees. We can now see that triangles $CDB$ and $CDA$ are both isosceles, with $CB=DB$ and $AD=AC$. From isosceles triangle $CDB$, we now know that $BD = 27$, and since $AB = c = 48$, we know that $AD = 21$. From isosceles triangle $CDA$, we now know that $CD = 21$. Applying Stewart's Theorem on triangle $ABC$ gives us $AC = 35$, which is $\boxed{\text{B}}$.

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