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Difference between revisions of "1974 AHSME Problems/Problem 23"

(Created page with "==Problem == In the adjoining figure <math> TP </math> and <math> T'Q </math> are parallel tangents to a circle of radius <math> r </math>, with <math> T </math> and <math> T' </...")
 
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==See Also==
 
==See Also==
 
{{AHSME box|year=1974|num-b=22|num-a=24}}
 
{{AHSME box|year=1974|num-b=22|num-a=24}}
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[[Category:Introductory Geometry Problems]]

Revision as of 09:20, 30 May 2012

Problem

In the adjoining figure $TP$ and $T'Q$ are parallel tangents to a circle of radius $r$, with $T$ and $T'$ the points of tangency. $PT''Q$ is a third tangent with $T'''$ as a point of tangency. If $TP=4$ and $T'Q=9$ then $r$ is

[asy] unitsize(45); pair O = (0,0); pair T = dir(90); pair T1 = dir(270); pair T2 = dir(25); pair P = (.61,1); pair Q = (1.61, -1); draw(unitcircle); dot(O); label("O",O,W); label("T",T,N); label("T'",T1,S); label("T''",T2,NE); label("P",P,NE); label("Q",Q,S); draw(O--T2); label("$r$",midpoint(O--T2),NW); draw(T--P); label("4",midpoint(T--P),N); draw(T1--Q); label("9",midpoint(T1--Q),S); draw(P--Q);[/asy]

$\mathrm{(A)\ } 25/6 \qquad \mathrm{(B) \ } 6 \qquad \mathrm{(C) \ } 25/4 \qquad$

$\mathrm{(D) \ } \text{a number other than }25/6, 6, 25/4 \qquad$

$\mathrm{(E) \ }\text{not determinable from the given information}$

Solution

[asy] unitsize(45); pair O = (0,0); pair T = dir(90); pair T1 = dir(270); pair T2 = dir(25); pair P = (.61,1); pair Q = (1.61, -1); draw(unitcircle); dot(O); label("O",O,W); label("T",T,N); label("T'",T1,S); label("T''",T2,NE); label("P",P,NE); label("Q",Q,S); draw(T--P); label("4",midpoint(T--P),N); draw(T1--Q); draw(P--Q); draw(T--T1); pair R = (.61,-1); draw(P--R); label("R",R,S); label("4",midpoint(T1--R),S); label("5",midpoint(R--Q),S); label("r",midpoint(O--T),W); label("r",midpoint(O--T1),W);[/asy]

Drop the perpendicular from $P$ to $T'Q$ and let the foot be $R$. Note that $PTT'R$ is a rectangle. Also, from Two Tangents, $PT''=4$ and $QT''=9$, so $PQ=13$. Therefore, from the Pythagorean Theorem on $\triangle PRQ$, $PR=\sqrt{13^2-5^2}=12$. We now see that $r=\frac{PR}{2}=6, \boxed{\text{B}}$.

See Also

1974 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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
All AHSME Problems and Solutions
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