1955 AHSME Problems/Problem 29

Problem

In the figure, $PA$ is tangent to semicircle $SAR$; $PB$ is tangent to semicircle $RBT$; $SRT$ is a straight line; the arcs are indicated in the figure. $\angle APB$ is measured by:

[asy] unitsize(1.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair O1=(0,0), O2=(3,0), Sp=(-2,0), R=(2,0), T=(4,0); pair A=O1+2*dir(60), B=O2+dir(85); pair Pa=rotate(90,A)*O1, Pb=rotate(-90,B)*O2; pair P=extension(A,Pa,B,Pb); pair[] dots={Sp,R,T,A,B,P}; draw(P--P+5*(A-P)); draw(P--P+5*(B-P)); clip((-2,0)--(-2,2.5)--(4,2.5)--(4,0)--cycle); draw(Arc(O1,2,0,180)--cycle); draw(Arc(O2,1,0,180)--cycle); dot(dots); label("$S$",Sp,S); label("$R$",R,S); label("$T$",T,S); label("$A$",A,NE); label("$B$",B,N); label("$P$",P,NNE); label("$a$",midpoint(Arc(O1,2,0,60)),SW); label("$b$",midpoint(Arc(O2,1,85,180)),SE); label("$c$",midpoint(Arc(O1,2,60,180)),SE); label("$d$",midpoint(Arc(O2,1,0,85)),SW);[/asy]

$\textbf{(A)}\ \frac{1}{2}(a-b)\qquad\textbf{(B)}\ \frac{1}{2}(a+b)\qquad\textbf{(C)}\ (c-a)-(d-b)\qquad\textbf{(D)}\ a-b\qquad\textbf{(E)}\ a+b$


Solution

[asy] unitsize(1.2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=3; pair O1=(0,0), O2=(3,0), Sp=(-2,0), R=(2,0), T=(4,0); pair A=O1+2*dir(60), B=O2+dir(85); pair Pa=rotate(90,A)*O1, Pb=rotate(-90,B)*O2; pair P=extension(A,Pa,B,Pb); pair[] dots={Sp,R,T,A,B,P}; draw(P--P+5*(A-P)); draw(P--P+5*(B-P)); clip((-2,0)--(-2,2.5)--(4,2.5)--(4,0)--cycle); draw(Arc(O1,2,0,180)--cycle); draw(Arc(O2,1,0,180)--cycle); draw(P--R, linetype("8 8")); draw(A--O1, linetype("8 8")); draw(B--O2, linetype("8 8")); draw(P--O1, linetype("8 8")); draw(P--O2, linetype("8 8")); dot(dots); label("$S$",Sp,S); label("$R$",R,S); label("$T$",T,S); label("$A$",A,NE); label("$B$",B,N); label("$P$",P,NNE); label("$a$",midpoint(Arc(O1,2,0,60)),SW); label("$b$",midpoint(Arc(O2,1,85,180)),SE); label("$c$",midpoint(Arc(O1,2,60,180)),SE); label("$d$",midpoint(Arc(O2,1,0,85)),SW);[/asy]

1955 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 28
Followed by
Problem 30
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