Difference between revisions of "1990 AHSME Problems/Problem 28"
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Or, after cross multiplying and writing in terms of the variables, <cmath>n^2-40n-r^2=0</cmath> | Or, after cross multiplying and writing in terms of the variables, <cmath>n^2-40n-r^2=0</cmath> | ||
Plugging in the value of <math>r</math> and solving the quadratic gives <math>n=CZ=71.5</math>, and from there we compute the desired difference to get <math>\fbox{(B) 13}</math>. | Plugging in the value of <math>r</math> and solving the quadratic gives <math>n=CZ=71.5</math>, and from there we compute the desired difference to get <math>\fbox{(B) 13}</math>. | ||
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== See also == | == See also == |
Latest revision as of 10:51, 5 October 2019
Problem
A quadrilateral that has consecutive sides of lengths and is inscribed in a circle and also has a circle inscribed in it. The point of tangency of the inscribed circle to the side of length 130 divides that side into segments of length and . Find .
Solution
Let , , , and be the vertices of this quadrilateral such that , , , and . Let be the center of the incircle. Draw in the radii from the center of the incircle to the points of tangency. Let these points of tangency , , , and be on , , , and , respectively. Using the right angles and the fact that the is cyclic, we see that quadrilaterals and are similar.
Let have length . Chasing lengths, we find that . Using Brahmagupta's Formula we find that has area and from that we find, using that fact that , where is the inradius and is the semiperimeter, .
From the similarity we have Or, after cross multiplying and writing in terms of the variables, Plugging in the value of and solving the quadratic gives , and from there we compute the desired difference to get .
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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.