2002 AIME I Problems/Problem 10
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 1
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.
Solution 2 Bash
By the Pythagorean Theorem, . From now on, every number we get will be rounded to two decimal points. Using law of cosines on AEF, EF is around 9.46. Using the angle bisector theorem, GF and DC are 7.28 and (185/7), respectively. We also know the lengths of EG and BD, so a quick Stewart's Theorem for triangles ABC and AEF gives lengths 3.76 and 14.75 for AG and AD, respectively. Now we have all segments of triangles AGF and ADC. Joy! It's time for some Heron's Formula. This gives area 10.95 for triangle AGF and 158.68 for triangle ADC. Subtracting gives us 147.73. Round to the nearest whole number and we get .
-jackshi2006
Solution 3 Coordinate Bash
By the Pythagorean Theorem, . By the Angle Bisector Theorem and . We can find the coordinates of F, and use that the find the equation of line EF. Then, we can find the coordinates of G. Triangle and we can find the area triangle with the shoelace theorem, so subtracting that from gives us as the closest integer.
-jackshi2006
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 |
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