Math contest question

[Uncle6]

Basically the question states to find the radius of the circle. A to N is 15, P to N is 9. I can already solve the problem using my method, solving for triangle OPQ and using Pythagorean theorem. However, in the results of 2007 math open challenge, they state that the most clever solution that they saw was where one extended OA to form the diameter OB and solved for that triangle. How do u know that triangle ABP is a similar triangle to APN?

source: http://www.math.ca/Competitions/COMC/
questions: http://www.math.ca/Competitions/COMC/COMC07/exame2007.pdf
solutions: http://www.math.ca/Competitions/COMC/COMC07/solutionse2007.pdf
results: http://www.math.ca/Competitions/COMC/COMC07/results2007.pdf
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Edited by stapel -- Reason for edit: Correcting link to graphic, providing links to other required information.

 

[mmm4444bot]

I cannot see the image.

(I do not currently have a PDF reader, so I cannot search through the 2007 problems.)

 

[Uncle6]

updated. Basically, I am ignorant in terms of triangle geometry. How to figure out similar trianlge?

 

[Denis]

On similar triangles: http://regentsprep.org/regents/math/sim ... rategy.htm

I have no idea why "they" think that extending lines and forming additional triangles is "clever",
when the solution is apparent...as you did it: r^2 = (r - 9)^2 + 15^2 ....r = 17

 

[mmm4444bot]

AP is a diagonal transversing QP and AN (which are parallel).

Therefore, angles AQP and PNA are right angles.

Also,

NAP = QPA

PNA = AQP

Since angles BAP and PAQ are the same angle,

and since angle BAP is part of a right triangle,

angle ABP must equal angle QPA.

This tells us that the interior angles of triangles ABP and APN are equal.

Two right triangles with equal interior angles are similar.