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 (INSANE BASH)
This solution is by no means one you should use during a competition. Attempt if you have a spare five hours.
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 .
I glossed over parts where I deduced irrational numbers and large fractions, but those shouldn't take too much time.
-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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