Difference between revisions of "2009 AIME II Problems/Problem 10"
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size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30)); | size(120); pathpen = linewidth(0.7); pointpen = black; pen f = fontsize(10); pair B=(0,0), A=(5,0), C=(0,13), E=(-5,0), O = incenter(E,C,A), D=IP(A -- A+3*(O-A),E--C); D(A--B--C--cycle); D(A--D--C); D(D--E--B, linetype("4 4")); MP("5", (A+B)/2, f); MP("13", (A+C)/2, NE,f); MP("A",D(A),f); MP("B",D(B),f); MP("C",D(C),N,f); MP("A'",D(E),f); MP("D",D(D),NW,f); D(rightanglemark(C,B,A,20)); D(anglemark(D,A,E,35));D(anglemark(C,A,D,30)); | ||
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== Solution 1== | == Solution 1== |
Latest revision as of 14:20, 19 August 2019
Problem
Four lighthouses are located at points , , , and . The lighthouse at is kilometers from the lighthouse at , the lighthouse at is kilometers from the lighthouse at , and the lighthouse at is kilometers from the lighthouse at . To an observer at , the angle determined by the lights at and and the angle determined by the lights at and are equal. To an observer at , the angle determined by the lights at and and the angle determined by the lights at and are equal. The number of kilometers from to is given by , where , , and are relatively prime positive integers, and is not divisible by the square of any prime. Find + + .
Diagram
-asjpz
Solution 1
Let be the intersection of and . By the Angle Bisector Theorem, = , so = and = , and + = = , so = , and = . Let be the foot of the altitude from to . It can be seen that triangle is similar to triangle , and triangle is similar to triangle . If = , then = , = , and = . Since + = = , = , and = (by the pythagorean theorem on triangle we sum and ). The answer is + + = .
Solution 2
Extend and to intersect at . Note that since and by ASA congruency we have . Therefore .
By the angle bisector theorem, and . Now we apply Stewart's theorem to find :
and our final answer is .
Solution 3
Notice that by extending and to meet at a point , is isosceles. Now we can do a straightforward coordinate bash. Let , , and , and the equation of line is . Let F be the intersection point of and , and by using the Angle Bisector Theorem: we have . Then the equation of the line through the points and is . Hence the intersection point of and is the point at the coordinates . Using the distance formula, for an answer of .
Solution 4
After drawing a good diagram, we reflect over the line , forming a new point that we'll call . Also, let the intersection of and be point . Point lies on line . Since line bisects , we can use the Angle Bisector Theorem. and , so . Letting the segments be and respectively, we now have . Therefore, . By the Pythagorean Theorem, . Using the Angle Bisector Theorem on , we have that . Substituting in , we have that , so the answer is .
(Solution by RootThreeOverTwo)
See Also
2009 AIME II (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.