Difference between revisions of "2002 AIME I Problems/Problem 10"
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== Solution == | == Solution == | ||
− | By the Pythagorean Theorem, <math>BC=35</math>. Letting <math>BD=x</math> we can use the | + | By the Pythagorean Theorem, <math>BC=35</math>. Letting <math>BD=x</math> we can use the Angle Bisector Theorem on triangle <math>ABC</math> to get <math>x/12=(35-x)/37</math>, and solving gives <math>BD=60/7</math> and <math>DC=185/7</math>. |
The area of triangle <math>AGF</math> is <math>10/3</math> that of triangle <math>AEG</math>, since they share a common side and angle, so the area of triangle <math>AGF</math> is <math>10/13</math> the area of triangle <math>AEF</math>. | The area of triangle <math>AGF</math> is <math>10/3</math> that of triangle <math>AEG</math>, since they share a common side and angle, so the area of triangle <math>AGF</math> is <math>10/13</math> the area of triangle <math>AEF</math>. |
Revision as of 22:35, 26 July 2016
Problem
In the diagram below, angle is a right angle. Point is on , and bisects angle . Points and are on and , respectively, so that and . Given that and , find the integer closest to the area of quadrilateral .
Solution
By the Pythagorean Theorem, . Letting we can use the Angle Bisector Theorem on triangle to get , and solving gives and .
The area of triangle is that of triangle , since they share a common side and angle, so the area of triangle is the area of triangle .
Since the area of a triangle is , the area of is and the area of is .
The area of triangle is , and the area of the entire triangle is . Subtracting the areas of and from and finding the closest integer gives as the answer.
See also
2002 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.