Difference between revisions of "2005 AIME II Problems/Problem 14"
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<cmath>\frac{AC}{\sin B}=\frac{AB}{\sin C}</cmath> | <cmath>\frac{AC}{\sin B}=\frac{AB}{\sin C}</cmath> | ||
Solving gets <math>\sin B=\frac{56}{65}</math>. | Solving gets <math>\sin B=\frac{56}{65}</math>. | ||
+ | |||
+ | Applying [[Stewart's Theorem]] with [[cevian]] <math>AE</math> we have | ||
+ | <cmath>(BC)(AE)^2+(BC)(BE)(CE)=(CE)(AB)^2+(BE)(AC)^2</cmath> | ||
+ | <cmath>(BC)(AE)^2+(BC)(BE)(BC-BE)=(BC-BE)(AB)^2+(BE)(AC)^2</cmath> | ||
+ | Solving gets <math>AE=\sqrt{\frac{15BE^2-198BE+2535}{15}</math> | ||
+ | |||
+ | Finally, applying [[Law of Sines]] on <math>\bigtriangle BAE</math> we have | ||
+ | <cmath>\frac{AE}{\sin B}=\frac{BE}{\sin BAE}</cmath> | ||
+ | <cmath>\frac{\sqrt{\frac{15BE^2-198BE+2535}{15}}{\frac{56}{65}}=\frac{BE}{\frac{6\sqrt{205}}{205}}</cmath> | ||
+ | <cmath>7605BE^2-32342BE+2535=0</cmath> | ||
+ | Solving | ||
== See also == | == See also == |
Revision as of 12:36, 22 March 2020
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2 (Similar Triangles)
- 4 Solution 3 (LoC and LoS bash)
- 5 Solution 4 (Ratio Lemma and Angle Bisector Theorem)
- 6 Solution 5 (Isogonal lines with respect to A angle bisesector)
- 7 Solution 6 (Tangent subtraction formulas)
- 8 Solution 7 (Super fast solution, 2 billion IQ)
- 9 Solution 8(literally 2 minute solution)
- 10 Solution 9 (Long but no IQ Required Altogether Bash)
- 11 See also
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,
Solution 5 (Isogonal lines with respect to A angle bisesector)
Since and are isogonal with respect to the angle bisector, we have To prove this, let and Then, by the Ratio Lemma, we have and multiplying these together proves the formula for isogonal lines. Hence, we have so our desired answer is
Solution 6 (Tangent subtraction formulas)
Note: We first recall some helpful tips regarding 13, 14, 15 triangles. Drawing an altitude H from B to AC results in AHB being a 5-12-13 right triangle and CHB being a 3-4-5 (9-12-15) right triangle.
Now we start by drawing altitudes from D and E onto AC, labeling them as F and G, and labelling . Now we know that and . Therefore, , so . Our goal now is to use tangent in triangle . We set to , so and , so and so . Now we just need tangent of .
We find this using , which is or . Now we solve the equation , so
Solution 7 (Super fast solution, 2 billion IQ)
Let
Via ratio lemma, we have and
Multiplying the two equations, we have . Plugging in the values we know, we have . Solving for , we get
Solution by hiker.
Solution 8(literally 2 minute solution)
Let , such that . Since and are isogonal, we get , and we can solve to get (and ). Hence, our answer is . - Spacesam
Solution 9 (Long but no IQ Required Altogether Bash)
Diagram borrowed from Solution 1.
Applying Law of Cosines on with respect to we have Solving gets , which implies that Applying Stewart's Theorem with cevian we have Solving gets .
Applying Law of Sines on to solve for we have Solving gets . Thus .
Applying Law of Sines on we have Solving gets .
Applying Stewart's Theorem with cevian we have Solving gets $AE=\sqrt{\frac{15BE^2-198BE+2535}{15}$ (Error compiling LaTeX. ! Missing } inserted.)
Finally, applying Law of Sines on $\bigtriangle BAE$ (Error compiling LaTeX. ! Undefined control sequence.) we have
\[\frac{\sqrt{\frac{15BE^2-198BE+2535}{15}}{\frac{56}{65}}=\frac{BE}{\frac{6\sqrt{205}}{205}}\] (Error compiling LaTeX. ! File ended while scanning use of \frac .)
Solving
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.