Difference between revisions of "2012 AIME I Problems/Problem 12"
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+ | ===Solution 7=== | ||
+ | WLOG, let <math>DE=8</math> and <math>BE=15</math>. First, by the Law of Sines on <math>\triangle CEB</math>, we find that | ||
+ | <cmath>\frac{\sin\angle B}{CE}=\frac{\sin\angle ECB}{BE}=\frac{1}{30}\implies \sin\angle B=\frac{CE}{30}.</cmath> | ||
+ | Now, we will find <math>CE</math>. Consider the following diagram: | ||
+ | [[Image:2012 AIME I -12 Sol 7 Diagram.png|200px|thumb|left| ]] | ||
+ | We have constructed equilateral triangle <math>\triangle BDP</math>, and its circumcircle. Since <math>\angle DCB=\angle DPB=60^\circ</math>, <math>C</math> lies on <math>(BDP)</math> as well. Let <math>Q</math> be the point diametrically opposite <math>P</math> on <math>(BCD)</math>, and let <math>R</math> be the foot of <math>Q</math> on <math>BD</math> (this is the midpoint of <math>BD</math>). It is easy to compute that <math>RQ=\frac{13}{\sqrt3}</math> and <math>ER=\frac{23}{2}-8=\frac{7}{2}</math>. Therefore, by the Pythagorean Theorem, <math>ER=\frac{169}{3}</math>. Now, by Power of a Point, we know that <math>(DE)(BE)=(EQ)(EC)</math>, which means that | ||
+ | <cmath>120=\frac{13EC}{\sqrt{3}}\implies EC=\frac{120\sqrt{3}}{13}.</cmath> | ||
+ | From before, we know that <math>\sin\angle B=\frac{EC}{30}\implies \sin\angle B=\frac{4\sqrt3}{13}</math>. It's now easy to compute <math>\cos\angle B</math> as well using the Pythagorean identity; we find that <math>\cos\angle B=\frac{11}{13}</math>, and thus <math>\tan\angle B=\frac{4\sqrt3}{11}</math> for an answer of <math>\boxed{18}</math>. | ||
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== See also == | == See also == | ||
{{AIME box|year=2012|n=I|num-b=11|num-a=13}} | {{AIME box|year=2012|n=I|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 02:13, 3 June 2020
Contents
Problem 12
Let be a right triangle with right angle at Let and be points on with between and such that and trisect If then can be written as where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. Find
Solutions
Solution 1
Let . Using angle bisector theorem on , , so . Then, drop the altitude from to and call the foot . Thus, , , and . Finally, . Our answer is .
Solution 2
Without loss of generality, set . Then, by the Angle Bisector Theorem on triangle , we have . We apply the Law of Cosines to triangle to get , which we can simplify to get .
Now, we have by another application of the Law of Cosines to triangle , so . In addition, , so .
Our final answer is .
Solution 3
(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)
Find values for all angles in terms of . , , , , and .
Use the law of sines on and :
In , . This simplifies to .
In , . This simplifies to .
Solve for and equate them so that you get .
From this, .
Use a trig identity on the denominator on the right to obtain:
This simplifies to
This gives Dividing by , we have
. Our final answer is .
Solution 4
(This solution avoids advanced trigonometry)
Let be the foot of the perpendicular from to , and let be the foot of the perpendicular from to .
Now let . Clearly, triangles and are similar with , so .
Since triangles and are 30-60-90 right triangles, we can easily find other lengths in terms of . For example, we see that and . Therefore .
Again using the fact that triangles and are similar, we see that , so .
Thus , and our answer is .
Solution 5
(Another solution without trigonometry)
Extend to point such that . It is then clear that is similar to .
Let , . Then .
With the Angle Bisector Theorem, we get that . From 30-60-90 , we get that and .
From , we have that . Simplifying yields , and , so our answer is .
Solution 6
Let , and let the feet of the altitudes from and to be and , respectively. Also, let and . We see that and by right triangles and . From this we have that . With the same triangles we have and . From 30-60-90 triangles and , we see that and , so . From our two values of we get:
Our answer is then .
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2012aimei/352
~ dolphin7
Solution 7
WLOG, let and . First, by the Law of Sines on , we find that Now, we will find . Consider the following diagram:
We have constructed equilateral triangle , and its circumcircle. Since , lies on as well. Let be the point diametrically opposite on , and let be the foot of on (this is the midpoint of ). It is easy to compute that and . Therefore, by the Pythagorean Theorem, . Now, by Power of a Point, we know that , which means that From before, we know that . It's now easy to compute as well using the Pythagorean identity; we find that , and thus for an answer of .
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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.