Difference between revisions of "1954 AHSME Problems/Problem 37"
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\textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad | \textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad | ||
\textbf{(E)}\ \text{none of these is correct} </math> | \textbf{(E)}\ \text{none of these is correct} </math> | ||
+ | |||
+ | == Solution == | ||
+ | Let <math>\angle RPQ</math> be <math>\theta</math>. | ||
+ | |||
+ | <math>p+ q + 2\theta = 180</math> | ||
+ | <math>m+\theta+90=180 \implies m+\theta=90 \implies 2m+2\theta=180</math> | ||
+ | <math>p+q+2\theta=2m+2\theta \implies \frac{p+q}{2}=m \implies \boxed{\textbf{(B) \ } \angle m = \frac{1}{2}(\angle p + \angle q)}</math> | ||
== Partial Solution == | == Partial Solution == | ||
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label("$D$",D,S,f);</asy> | label("$D$",D,S,f);</asy> | ||
Looking at triangle PRQ, we have <math>\angle RPD+\angle RQS+\angle MRN=180</math> and from the given statement <math>\angle NMR=\frac{1}{2}\angle MRN</math>, so looking at triangle MOR <math>\angle NMR=90-\frac{\angle RPD+\angle RQS}{2}</math>, which rules out | Looking at triangle PRQ, we have <math>\angle RPD+\angle RQS+\angle MRN=180</math> and from the given statement <math>\angle NMR=\frac{1}{2}\angle MRN</math>, so looking at triangle MOR <math>\angle NMR=90-\frac{\angle RPD+\angle RQS}{2}</math>, which rules out | ||
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+ | |||
+ | {{AHSME 50p box|year=1954|num-b=37|num-a=39}} | ||
+ | {{MAA Notice}} |
Revision as of 16:22, 15 April 2017
Problem 37
Given with bisecting , extended to and a right angle, then:
Solution
Let be .
Partial Solution
Looking at triangle PRQ, we have and from the given statement , so looking at triangle MOR , which rules out
1954 AHSC (Problems • Answer Key • Resources) | ||
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Followed by Problem 39 | |
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