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Difference between revisions of "1954 AHSME Problems/Problem 37"

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<math>\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad  
 
<math>\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad  
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\textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q) </math>\qquad
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\textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q) \qquad
−
<math> \textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad
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\textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad
 
\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>
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== Solution ==
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Let <math>\angle PRS</math> be <math>\theta</math>.
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<math>p+ q + 2\theta = 180</math>
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<math>m+\theta+90=180 \implies m+\theta=90 \implies 2m+2\theta=180</math>
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<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=36|num-a=38}}
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{{MAA Notice}}

Latest revision as of 09:11, 3 May 2020

Problem 37

Given $\triangle PQR$ with $\overline{RS}$ bisecting $\angle R$, $PQ$ extended to $D$ and $\angle n$ a right angle, then:

[asy] path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false) { pair M,N; path mark; M=t*0.03*unit(A-B)+B; N=t*0.03*unit(C-B)+B; if(flip) mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B)); else mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B)); return mark; } unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair P=(0,0), R=(3,2), Q=(4,0); pair S0=bisectorpoint(P,R,Q); pair Sp=extension(P,Q,S0,R); pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp); pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R); draw(P--R--Q); draw(R--Sp); draw(P--D--M); draw(anglemark2(Sp,P,R,17)); label("$p$",P+(0.35,0.1)); draw(anglemark2(R,Q,P,11)); label("$q$",Q+(-0.17,0.1)); draw(anglemark2(R,Np,D,8,true)); label("$n$",Np+(+0.12,0.07)); draw(anglemark2(R,M,D,13,true)); label("$m$",M+(+0.25,0.03)); draw(anglemark2(M,D,P,29)); label("$d$",D+(-0.75,0.095)); pen f=fontsize(10pt); label("$R$",R,N,f); label("$P$",P,S,f); label("$S$",Sp,S,f); label("$Q$",Q,S,f); label("$D$",D,S,f);[/asy]

$\textbf{(A)}\ \angle m = \frac {1}{2}(\angle p - \angle q) \qquad \textbf{(B)}\ \angle m = \frac {1}{2}(\angle p + \angle q) \qquad \textbf{(C)}\ \angle d =\frac{1}{2}(\angle q+\angle p)\qquad \textbf{(D)}\ \angle d =\frac{1}{2}\angle m\qquad \textbf{(E)}\ \text{none of these is correct}$

Solution

Let $\angle PRS$ be $\theta$.

$p+ q + 2\theta = 180$

$m+\theta+90=180 \implies m+\theta=90 \implies 2m+2\theta=180$

$p+q+2\theta=2m+2\theta \implies \frac{p+q}{2}=m \implies \boxed{\textbf{(B) \ } \angle m = \frac{1}{2}(\angle p + \angle q)}$

Partial Solution

[asy] import math; path anglemark2(pair A, pair B, pair C, real t=8, bool flip=false) { pair M,N; path mark; M=t*0.03*unit(A-B)+B; N=t*0.03*unit(C-B)+B; if(flip) mark=Arc(B,t*0.03,degrees(C-B)-360,degrees(A-B)); else mark=Arc(B,t*0.03,degrees(A-B),degrees(C-B)); return mark; } unitsize(1.5cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); pair P=(0,0), R=(3,2), Q=(4,0); pair S0=bisectorpoint(P,R,Q); pair Sp=extension(P,Q,S0,R); pair D0=bisectorpoint(R,Sp), Np=midpoint(R--Sp); pair D=extension(Np,D0,P,Q), M=extension(Np,D0,P,R); draw(P--R--Q); draw(R--Sp); draw(P--D--M); pen f=fontsize(10pt); pair pI=extension(D,M,R,Q); label("$O$",pI+(-0.2,0.166),f); draw(anglemark2(Sp,P,R,17)); label("$p$",P+(0.35,0.1)); draw(anglemark2(R,Q,P,11)); label("$q$",Q+(-0.17,0.1)); draw(anglemark2(R,Np,D,8,true)); label("$n$",Np+(+0.12,0.07)); draw(anglemark2(R,M,D,13,true)); label("$m$",M+(+0.25,0.03)); draw(anglemark2(M,D,P,29)); label("$d$",D+(-0.75,0.095)); label("$R$",R,N,f); label("$M$",M+(-.07,.07),f); label("$N$",Np+(-.08,.15),f); label("$P$",P,S,f); label("$S$",Sp,S,f); label("$Q$",Q,S,f); label("$D$",D,S,f);[/asy] Looking at triangle PRQ, we have $\angle RPD+\angle RQS+\angle MRN=180$ and from the given statement $\angle NMR=\frac{1}{2}\angle MRN$, so looking at triangle MOR $\angle NMR=90-\frac{\angle RPD+\angle RQS}{2}$, which rules out


1954 AHSC (Problems • Answer Key • Resources)
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