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 ~ \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 \textbf{(B)} | + | 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 ~ \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>\textbf{(B)}</math> |
==See Also== | ==See Also== |
Revision as of 18:43, 20 October 2018
Problem 18
Circle has diameters
and
perpendicular to each other.
is any chord intersecting
at
.
Then
is equal to:
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 |
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