Difference between revisions of "2002 AIME I Problems/Problem 10"
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By the Pythagorean Theorem, <math>BC=35</math>. 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 <math>\boxed{148}</math>. | By the Pythagorean Theorem, <math>BC=35</math>. 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 <math>\boxed{148}</math>. | ||
Revision as of 16:21, 5 December 2020
Contents
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
.
![AIME 2002I Problem 10.png](https://wiki-images.artofproblemsolving.com//8/8d/AIME_2002I_Problem_10.png)
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.