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Difference between revisions of "2002 AIME I Problems/Problem 10"

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== Solution 2 (INSANE BASH) ==
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== Solution 2 Bash==
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This solution is by no means one you should use during a competition. Attempt if you have a spare five hours.
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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>.
  
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-jackshi2006
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== Solution 3 Coordinate Bash ==
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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.
  
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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>.
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-jackshi2006
  
  
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I glossed over parts where I deduced irrational numbers and large fractions, but those shouldn't take too much time.
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== Solution 4 No Trig ==
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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>
  
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-jackshi2006
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~ 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 21:44, 8 August 2023

Problem

In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$, and $\overline{AD}$ bisects angle $CAB$. Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$, respectively, so that $AE=3$ and $AF=10$. Given that $EB=9$ and $FC=27$, find the integer closest to the area of quadrilateral $DCFG$.

AIME 2002I Problem 10.png

Solution 1

By the Pythagorean Theorem, $BC=35$. Letting $BD=x$ we can use the Angle Bisector Theorem on triangle $ABC$ to get $x/12=(35-x)/37$, and solving gives $BD=60/7$ and $DC=185/7$.

The area of triangle $AGF$ is $10/3$ that of triangle $AEG$, since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$.

Since the area of a triangle is $\frac{ab\sin{C}}2$, the area of $AEF$ is $525/37$ and the area of $AGF$ is $5250/481$.

The area of triangle $ABD$ is $360/7$, and the area of the entire triangle $ABC$ is $210$. Subtracting the areas of $ABD$ and $AGF$ from $210$ and finding the closest integer gives $\boxed{148}$ as the answer.


Solution 2 Bash

By the Pythagorean Theorem, $BC=35$. 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 $\boxed{148}$.

-jackshi2006

Solution 3 Coordinate Bash

By the Pythagorean Theorem, $BC=35$. By the Angle Bisector Theorem $BD = 60/7$ and $DC = 185/7$. 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 $ADC = 1110/7$ and we can find the area triangle $AGF$ with the shoelace theorem, so subtracting that from $ADC$ gives us $\boxed{148}$ as the closest integer.

-jackshi2006


Solution 4 No Trig

By the Pythagorean Theorem, $BC = 35$, and by the Angle Bisector Theorem $BD = 60/7$ and $DC = 185/7$. Draw a perpendicular from $F$ to $\overline{AE}$. Let the intersection of $F$ and $\overline{AE}$ be $H$. triangle $AHF$ is similar to $ABC$ by $AA$ similarity. thus, $AF/AC = HF/BC$. We find that $HF = 350/37$, so the area of $AEF = 525/37$. The area of triangle $AGF$ is $10/3$ that of triangle $AEG$, since they share a common side and angle, so the area of triangle $AGF$ is $10/13$ the area of triangle $AEF$, so the area of $AGF = 5250/481$. The area of triangle $ADC$ is $1110/7$, since the base is $185/7$ and the height is $12$. Thus, the area of $DCFG$ equals the area of $ADC - AGF$, or rounded to the nearest integer, $\boxed{148}$

~ 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

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