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== |
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.