1961 AHSME Problems/Problem 25

Problem

$\triangle ABC$ is isosceles with base $AC$ . Points $P$ and $Q$ are respectively in $CB$ and $AB$ and such that $AC=AP=PQ=QB$ . The number of degrees in $\angle B$ is:

$\textbf{(A)}\ 25\frac{5}{7}\qquad \textbf{(B)}\ 26\frac{1}{3}\qquad \textbf{(C)}\ 30\qquad \textbf{(D)}\ 40\qquad \textbf{(E)}\ \text{Not determined by the information given}$

Solution

Let $b$ be the measure of $\angle B$ . $\triangle BQP$ is an isosceles triangle, so $\angle BPQ = b$ and $\angle BQP = 180 - 2b$ .

$AB$ is a line, so $\angle AQP = 2b$ . Since $\triangle QPA$ is isosceles as well, $\angle QAP = 2b$ and $\angle QPA = 180 - 4b$ .

$BC$ is a line, so $\angle APC = 3b$ . Since $\triangle APC$ is isosceles as well, $\angle ACB = 3b$ . $\triangle ABC$ is also isosceles, so $\angle BAC = 3b$ , so $\angle PAC = b$ .

The angles in a triangle add up to $180$ degrees, so $3b + 3b + b = 180$ . Solving the equation yields $b = \frac{180}{7}$ . Thus, $\angle B = 25\frac{5}{7}^{\circ}$ , so the answer is $\boxed{\textbf{(A)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1961 AHSME Problems/Problem 25

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