Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 9"
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Let AB=x. Call the foot of the perpendicular from D to AB N, and the foot of the perpendicular from C to AB M. By similarity, AN=2x/17. Also, AM=x/2. Since <math>\triangle</math>AND and <math>\triangle</math>CAM are similar, we have (2x/17)/AD=(x/2)/16. Hence, AD=64/17, and CD=16-AD=208/17, so the answer is 225. | Let AB=x. Call the foot of the perpendicular from D to AB N, and the foot of the perpendicular from C to AB M. By similarity, AN=2x/17. Also, AM=x/2. Since <math>\triangle</math>AND and <math>\triangle</math>CAM are similar, we have (2x/17)/AD=(x/2)/16. Hence, AD=64/17, and CD=16-AD=208/17, so the answer is 225. | ||
| − | ==See | + | ==See Also== |
| + | {{Mock AIME box|year=Pre 2005|n=3|num-b=8|num-a=10}} | ||
Revision as of 10:36, 4 April 2012
Problem
is an isosceles triangle with base 
. 
is a point on 
and 
is the point on the extension of 
past such that 
is right. If 
and 
, then 
can be expressed as 
, where 
and 
are relatively prime positive integers. Determine 
.
Solution
Let AB=x. Call the foot of the perpendicular from D to AB N, and the foot of the perpendicular from C to AB M. By similarity, AN=2x/17. Also, AM=x/2. Since 
AND and CAM are similar, we have (2x/17)/AD=(x/2)/16. Hence, AD=64/17, and CD=16-AD=208/17, so the answer is 225.
See Also
| Mock AIME 3 Pre 2005 (Problems, Source) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
