Difference between revisions of "1957 AHSME Problems/Problem 18"
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− | == Problem | + | == Problem == |
Circle <math>O</math> has diameters <math>AB</math> and <math>CD</math> perpendicular to each other. <math>AM</math> is any chord intersecting <math>CD</math> at <math>P</math>. | Circle <math>O</math> has diameters <math>AB</math> and <math>CD</math> perpendicular to each other. <math>AM</math> is any chord intersecting <math>CD</math> at <math>P</math>. | ||
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− | ==Solution== | + | == Solution 1 == |
− | Draw <math>MB</math>. Since <math>\angle AMB</math> is inscribed on a diameter, <math>\angle AMB</math> is <math>90^\circ</math>. By AA Similarity, <math>\triangle APO | + | Draw <math>MB</math>. Since <math>\angle AMB</math> is inscribed on a diameter, <math>\angle AMB</math> is <math>90^\circ</math>. By [[AA Similarity]], <math>\triangle APO \sim \triangle ABM</math>. Setting up ratios, we get <math>\frac{AP}{AO}=\frac{AB}{AM}</math>. Cross-multiplying, we get <math>AP\cdot AM = AO \cdot AB</math>, so the answer is <math>\boxed{\textbf{(B)}}</math>. |
− | == | + | == Solution 2 == |
+ | By [[Thales' Theorem]], <math>\measuredangle PMB = \measuredangle AMB = 90^{\circ}</math>. Because, from the problem, <math>\measuredangle POB = 90^{\circ}</math> as well, <math>\measuredangle PMB + \measuredangle POB = 180^{\circ}</math>, so <math>PMBO</math> is a [[cyclic quadrilateral]]. Thus, because <math>P</math>, <math>M</math>, <math>O</math>, and <math>B</math> lie on a circle, we can use [[Power of a Point]]. From this theorem, we get that <math>AP \cdot AM = AO \cdot AB</math>, which is answer choice <math>\boxed{\textbf{(B)}}</math>. | ||
− | {{AHSME box|year=1957|num-b=17|num-a=19}} | + | |
+ | == See Also == | ||
+ | |||
+ | {{AHSME 50p box|year=1957|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | [[Category:Introductory Geometry Problems]] |
Latest revision as of 09:53, 25 July 2024
Contents
Problem
Circle has diameters
and
perpendicular to each other.
is any chord intersecting
at
.
Then
is equal to:
Solution 1
Draw . Since
is inscribed on a diameter,
is
. By AA Similarity,
. Setting up ratios, we get
. Cross-multiplying, we get
, so the answer is
.
Solution 2
By Thales' Theorem, . Because, from the problem,
as well,
, so
is a cyclic quadrilateral. Thus, because
,
,
, and
lie on a circle, we can use Power of a Point. From this theorem, we get that
, which is answer choice
.
See Also
1957 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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All AHSME Problems and Solutions |
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