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>. | ||
+ | -jackshi2006 | ||
− | + | == Solution 3 Coordinate Bash == | |
+ | By the Pythagorean Theorem, <math>BC=35</math>. By the Angle Bisector Theorem <math>BD = 60/7</math> and <math>DC = 185/7</math>. 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 <math>ADC = 1110/7</math> and we can find the area triangle <math>AGF</math> with the shoelace theorem, so subtracting that from <math>ADC</math> gives us <math>\boxed{148}</math> as the closest integer. | ||
-jackshi2006 | -jackshi2006 | ||
+ | |||
+ | |||
+ | == Solution 4 No Trig == | ||
+ | By the Pythagorean Theorem, <math>BC = 35</math>, and by the Angle Bisector Theorem <math>BD = 60/7</math> and <math>DC = 185/7</math>. Draw a perpendicular from <math>F</math> to <math>\overline{AE}</math>. Let the intersection of <math>F</math> and <math>\overline{AE}</math> be <math>H</math>. triangle <math>AHF</math> is similar to <math>ABC</math> by <math>AA</math> similarity. thus, <math>AF/AC = HF/BC</math>. We find that <math>HF = 350/37</math>, so the area of <math>AEF = 525/37</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>, so the area of <math>AGF = 5250/481</math>. The area of triangle <math>ADC</math> is <math>1110/7</math>, since the base is <math>185/7</math> and the height is <math>12</math>. Thus, the area of <math>DCFG</math> equals the area of <math>ADC - AGF</math>, or rounded to the nearest integer, <math>\boxed{148}</math> | ||
+ | |||
+ | ~ PaperMath | ||
== 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}} |
Latest revision as of 22:44, 8 August 2023
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
.
-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
Solution 4 No Trig
By the Pythagorean Theorem, , and by the Angle Bisector Theorem
and
. Draw a perpendicular from
to
. Let the intersection of
and
be
. triangle
is similar to
by
similarity. thus,
. We find that
, so the area of
. 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
, so the area of
. The area of triangle
is
, since the base is
and the height is
. Thus, the area of
equals the area of
, or rounded to the nearest integer,
~ PaperMath
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.