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Difference between revisions of "1961 AHSME Problems/Problem 25"

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Let <math>b</math> be the measure of <math>\angle B</math>.  <math>\triangle BQP</math> is an [[isosceles triangle]], so <math>\angle BPQ = b</math> and <math>\angle BQP = 180 - 2b</math>.
 
Let <math>b</math> be the measure of <math>\angle B</math>.  <math>\triangle BQP</math> is an [[isosceles triangle]], so <math>\angle BPQ = b</math> and <math>\angle BQP = 180 - 2b</math>.

Latest revision as of 11:07, 31 May 2018

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

1961 AHSME Problem 25.png

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 $\boxed{\textbf{(A)}}$.

See Also

1961 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Problem 26
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All AHSME Problems and Solutions

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