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Difference between revisions of "1960 AHSME Problems/Problem 32"

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\textbf{(D)}\ AB \times AD = OB \times AO\qquad \\
 
\textbf{(D)}\ AB \times AD = OB \times AO\qquad \\
 
\textbf{(E)}\ \text{none of these}    </math>
 
\textbf{(E)}\ \text{none of these}    </math>
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== Solution 1==
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We claim that <math>\fbox{A}</math> is the right answer.
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Let the radius of circle <math>O</math> be <math>r</math>, and let the length of <math>AD=x</math>. Since <math>AB \perp BC</math>, <math>AB</math> is a tangent to circle <math>O</math>. Thus, by the tangent-secant theorem, we have <math>AB^2=AD\times AE</math>, or, <math>(2r)^2=x(x+2r)</math>. Through some algebraic manipulation, we find <cmath>\begin{align}AD^2=x^2=2r(2r-x)\end{align}</cmath>.
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Since <math>AD^2=AP^2=x^2</math>, <math>PB=AB-AP=2r-x</math>, and <math>AB=2r</math>, we see that <math>(1)</math> is identical to <math>AP^2=PB\times AB</math>, hence our answer is <math>\fbox{A}</math>
  
 
==Solution 2 (guess and check) ==
 
==Solution 2 (guess and check) ==

Latest revision as of 19:33, 17 August 2022

Problem

In this figure the center of the circle is $O$. $AB \perp BC$, $ADOE$ is a straight line, $AP = AD$, and $AB$ has a length twice the radius. Then:

[asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); real e=350,c=55; pair O=origin,E=dir(e),C=dir(c),B=dir(180+c),D=dir(180+e), rot=rotate(90,B)*O,A=extension(E,D,B,rot); path tangent=A--B; pair P=waypoint(tangent,abs(A-D)/abs(A-B)); draw(unitcircle^^C--B--A--E); dot(A^^B^^C^^D^^E^^P,linewidth(2)); label("$O$",O,dir(290)); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,NE); label("$D$",D,dir(120)); label("$E$",E,SE); label("$P$",P,SW);[/asy]

$\textbf{(A)} AP^2 = PB \times AB\qquad \\ \textbf{(B)}\ AP \times DO = PB \times AD\qquad \\ \textbf{(C)}\ AB^2 = AD \times DE\qquad \\ \textbf{(D)}\ AB \times AD = OB \times AO\qquad \\ \textbf{(E)}\ \text{none of these}$

Solution 1

We claim that $\fbox{A}$ is the right answer.

Let the radius of circle $O$ be $r$, and let the length of $AD=x$. Since $AB \perp BC$, $AB$ is a tangent to circle $O$. Thus, by the tangent-secant theorem, we have $AB^2=AD\times AE$, or, $(2r)^2=x(x+2r)$. Through some algebraic manipulation, we find \begin{align}AD^2=x^2=2r(2r-x)\end{align}.

Since $AD^2=AP^2=x^2$, $PB=AB-AP=2r-x$, and $AB=2r$, we see that $(1)$ is identical to $AP^2=PB\times AB$, hence our answer is $\fbox{A}$

Solution 2 (guess and check)

Let $r$ be the radius of the circle, so $AB = 2r$. By the Pythagorean Theorem, $AO = \sqrt{(2r)^2 + r^2} = r \sqrt{5}$. That means, $AD = AP = r \sqrt{5} - r$, so $PB = 2r - r\sqrt{5} + r = 3r - r\sqrt{5}$.

Substitute values for each answer choice to determine which one is correct for all $r$.

For option A, substitution results in \[(r \sqrt{5} - r)^2 = (3r - r\sqrt{5})2r\] \[5r^2 - 2r^2 \sqrt{5} + r^2 = 6r^2 - 2r^2 \sqrt{5}\] \[6r^2 - 2r^2 \sqrt{5} = 6r^2 - 2r^2 \sqrt{5}\]

For option B, substitution results in \[(r \sqrt{5} - r)r = (3r - r\sqrt{5})(r \sqrt{5} - r)\] \[r^2 \sqrt{5} - r^2 = 4r^2 \sqrt{5} - 8r^2\]

For option C, substitution results in \[(2r)^2 = (r\sqrt{5} - r)2r\] \[4r^2 = 2r^2 \sqrt{5} - 2r^2\]

For option D, substitution results in \[(r \sqrt{5} - r) \cdot 2r = r \cdot r\sqrt{5}\] \[2r^2 \sqrt{5} - 2r^2 = r^2 \sqrt{5}\]

From each option, only option A has both sides equaling each other, so the answer is $\boxed{\textbf{(A)}}$.

See Also

1960 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 31 		Followed by
Problem 33
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
All AHSME Problems and Solutions
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