Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 53: Line 53:
  
 
<math>p+ q + 2\theta = 180</math>
 
<math>p+ q + 2\theta = 180</math>
 +
 
<math>m+\theta+90=180 \implies m+\theta=90 \implies 2m+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>
 
<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>
  

Revision as of 16:23, 15 April 2017

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)
Preceded by
Problem 37 		Followed by
Problem 39
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_37&oldid=85220"