Difference between revisions of "2005 AIME II Problems/Problem 14"
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Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively. | Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively. | ||
− | From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8}</math> | + | From here, we can use Heron's Formula to find the altitude. The area of the triangle is <math>\sqrt{21*6*7*8} = 84</math>. We can then use similar triangles with triangle <math>AQC</math> and triangle <math>DSC</math> to find <math>DS=\frac{24}{5}</math>. Consequently, from Pythagorean theorem, <math>SC = \frac{18}{5}</math> and <math>AS = 14-SC = \frac{52}{5}</math>. We can also use the Pythagorean theorem on triangle <math>AQB</math> to determine that <math>BQ = \frac{33}{5}</math>. |
− | Label AR as y and RE as x. RB then equals 13-y. Then, we have two similar triangles. | + | Label <math>AR</math> as <math>y</math> and <math>RE</math> as <math>x</math>. <math>RB</math> then equals <math>13-y</math>. Then, we have two similar triangles. |
Firstly: <math>\triangle ARE \sim \triangle ASD</math>. From there, we have <math>\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}</math>. | Firstly: <math>\triangle ARE \sim \triangle ASD</math>. From there, we have <math>\frac{x}{y} = \frac{\frac{24}{5}}{\frac{53}{5}}</math>. | ||
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Next: <math>\triangle BRE \sim \triangle BQA</math>. From there, we have <math>\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}</math>. | Next: <math>\triangle BRE \sim \triangle BQA</math>. From there, we have <math>\frac{x}{13-y} = \frac{\frac{56}{5}}{\frac{33}{5}}</math>. | ||
− | Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. Notice that 463 is prime, so even though we use | + | Solve the system to get <math>x = \frac{2184}{463}</math> and <math>y = \frac{4732}{463}</math>. Notice that 463 is prime, so even though we use the Pythagorean theorem on <math>x</math> and <math>13-y</math>, the denominator won't change. The answer we desire is <math>\boxed{463}</math>. |
==Solution 3 (LoC and LoS bash)== | ==Solution 3 (LoC and LoS bash)== |
Revision as of 18:02, 10 March 2018
Contents
Problem
In triangle and Point is on with Point is on such that Given that where and are relatively prime positive integers, find
Solution 1
By the Law of Sines and since , we have
Substituting our knowns, we have . The answer is .
Solution 2 (Similar Triangles)
Drop the altitude from A and call the base of the altitude Q. Also, drop the altitudes from E and D to AB and AC respectively. Call the feet of the altitudes R and S respectively.
From here, we can use Heron's Formula to find the altitude. The area of the triangle is . We can then use similar triangles with triangle and triangle to find . Consequently, from Pythagorean theorem, and . We can also use the Pythagorean theorem on triangle to determine that .
Label as and as . then equals . Then, we have two similar triangles.
Firstly: . From there, we have .
Next: . From there, we have .
Solve the system to get and . Notice that 463 is prime, so even though we use the Pythagorean theorem on and , the denominator won't change. The answer we desire is .
Solution 3 (LoC and LoS bash)
Let . Note by Law of Sines on we have As a result, our goal is to find and (we already know ).
Let the foot of the altitude from to be . By law of cosines on we have It follows that and .
Note that by PT on we have that . By Law of Sines on (where we square everything to avoid taking the square root) we see How are we going to find though? and are in the same triangle. Applying Law of Sines on we see that , , and are all in the same triangle. We know they add up to . There's a good chance we can exploit this using the identity .
We have that . Success! We know and already. Applying the addition formula we see This is the last stretch! Applying Law of Sines a final time on we see It follows that the answer is .
Solution 4 (Ratio Lemma and Angle Bisector Theorem)
Let be the angle bisector of such that is on .
Then , and thus .
By the Ratio Lemma, and .
This implies that .
Thus, .
. Thus, .
Additionally, . Solving gives that
Alternate: By the ratio lemma,
Combining these, we get
Thus,
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.