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 \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>. | 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 08:53, 25 July 2024
Contents
Problem
Circle 
has diameters 
and 
perpendicular to each other. 
is any chord intersecting at 
.
Then 
is equal to:
![[asy] defaultpen(linewidth(.8pt)); unitsize(2cm); pair O = origin; pair A = (-1,0); pair B = (1,0); pair C = (0,1); pair D = (0,-1); pair M = dir(45); pair P = intersectionpoint(O--C,A--M); draw(Circle(O,1)); draw(A--B); draw(C--D); draw(A--M); label("$A$",A,W); label("$B$",B,E); label("$C$",C,N); label("$D$",D,S); label("$M$",M,NE); label("$O$",O,NE); label("$P$",P,NW);[/asy]](http://latex.artofproblemsolving.com/a/3/d/a3da30e392e978e1b0205ac5cd8f006069965b7d.png)
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 | |
| 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. 
