Difference between revisions of "1999 AIME Problems/Problem 14"
m (→Solution 4) |
m (→Solution 4) |
||
Line 120: | Line 120: | ||
== Solution 4 == | == Solution 4 == | ||
− | By splitting the triangle down its altitude, we get a <math>5,12,13</math> and <math>9,12,15</math> triangle. Then <math>\cos\angle{ABC} = \frac{5}{13}</math> and <math>cos\angle{ACB} = \frac{3}{5}</math>. We do this similarly for <math>\cos\angle CAB</math> using Law of Cosine, this is <math>\frac{33}{65}</math>. Now use Law of Sine on triangle <math>PCA</math>. We find that <math>\frac{\sin{\alpha}}{AP} = \frac{\sin{\angle{CPA}}}{15}</math>. Substituting values in we get <math>\frac{\sin{\alpha}}{AP} = \frac{\frac{56}{65}}{15}</math>. Similar expression for triangle <math>BPA</math>, relating expression for <math>AP</math> on that side gets us 2 equation 2 unknown for <math>\alpha</math> and <math>AP</math>. Solving for the tangent of alpha, we get <math>\frac{168}{295} = \boxed{463}</math>. | + | By splitting the triangle down its altitude, we get a <math>5,12,13</math> and <math>9,12,15</math> triangle. Then <math>\cos\angle{ABC} = \frac{5}{13}</math> and <math>\cos\angle{ACB} = \frac{3}{5}</math>. We do this similarly for <math>\cos\angle CAB</math> using Law of Cosine, this is <math>\frac{33}{65}</math>. Now use Law of Sine on triangle <math>PCA</math>. We find that <math>\frac{\sin{\alpha}}{AP} = \frac{\sin{\angle{CPA}}}{15}</math>. Substituting values in we get <math>\frac{\sin{\alpha}}{AP} = \frac{\frac{56}{65}}{15}</math>. Similar expression for triangle <math>BPA</math>, relating expression for <math>AP</math> on that side gets us 2 equation 2 unknown for <math>\alpha</math> and <math>AP</math>. Solving for the tangent of alpha, we get <math>\frac{168}{295} = \boxed{463}</math>. |
== See also == | == See also == |
Revision as of 22:29, 28 May 2021
Problem
Point is located inside triangle so that angles and are all congruent. The sides of the triangle have lengths and and the tangent of angle is where and are relatively prime positive integers. Find
Solution
Solution 1
Drop perpendiculars from to the three sides of and let them meet and at and respectively.
Let and . We have that We can then use the tool of calculating area in two ways On the other hand, We still need though. We have all these right triangles and we haven't even touched Pythagoras. So we give it a shot: Adding gives Recall that we found that . Plugging in , we get , giving us for an answer.
Solution 2
Let , , , , , and .
So by the Law of Cosines, we have: Adding these equations and rearranging, we have: Now , by Heron's formula.
Now the area of a triangle, , where and are sides on either side of an angle, . So, Adding these equations yields: Dividing by , we have: Thus, .
Note: In fact, this problem is unfairly easy to those who happen to have learned about Brocard point. The Brocard Angle is given by
Solution 3
Let Then, using Law of Cosines on the three triangles containing vertex we have Add the three equations up and rearrange to obtain Also, using we have Divide the two equations to obtain
Solution 4
By splitting the triangle down its altitude, we get a and triangle. Then and . We do this similarly for using Law of Cosine, this is . Now use Law of Sine on triangle . We find that . Substituting values in we get . Similar expression for triangle , relating expression for on that side gets us 2 equation 2 unknown for and . Solving for the tangent of alpha, we get .
See also
1999 AIME (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.