Difference between revisions of "2002 AIME I Problems/Problem 10"
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==Solution 2 (Desperate Bash)== | ==Solution 2 (Desperate Bash)== | ||
− | We want to find the area of <math>DCFG</math>. This is the same as <math>[ADC]</math> - <math>[AGF]</math>. We know <math>[ADC]</math> is <math>\frac{ | + | We want to find the area of <math>DCFG</math>. This is the same as <math>[ADC]</math> - <math>[AGF]</math>. |
− | From the Pythagorean Theorem, < | + | |
+ | Let us calculate <math>[ADC]</math> first. We know <math>[ADC]</math> is <math>\frac{AD*AC\sin{A}}{2}</math>. Firstly, we need AD. From the angle bisector theorem | ||
+ | From the Pythagorean Theorem, <math>BC=35</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2002|n=I|num-b=9|num-a=11}} | {{AIME box|year=2002|n=I|num-b=9|num-a=11}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:09, 3 July 2015
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.
Solution 2 (Desperate Bash)
We want to find the area of . This is the same as - .
Let us calculate first. We know is . Firstly, we need AD. From the angle bisector theorem From the Pythagorean Theorem, .
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.