Difference between revisions of "1957 AHSME Problems/Problem 18"
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==Solution== | ==Solution== | ||
| − | 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>. |
==See Also== | ==See Also== | ||
Revision as of 08:41, 25 July 2024
Problem 18
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
Draw 
. Since 
is inscribed on a diameter, is 
. By AA Similarity, 
. Setting up ratios, we get 
. Cross-multiplying, we get 
, so the answer is 
.
See Also
| 1957 AHSME (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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
