Difference between revisions of "2009 AIME II Problems/Problem 10"
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− | Four lighthouses are located at points , , , and . The lighthouse at | + | ==Problem== |
+ | Four lighthouses are located at points <math>A</math>, <math>B</math>, <math>C</math>, and <math>D</math>. The lighthouse at <math>A</math> is <math>5</math> kilometers from the lighthouse at <math>B</math>, the lighthouse at <math>B</math> is <math>12</math> kilometers from the lighthouse at <math>C</math>, and the lighthouse at <math>A</math> is <math>13</math> kilometers from the lighthouse at <math>C</math>. To an observer at <math>A</math>, the angle determined by the lights at <math>B</math> and <math>D</math> and the angle determined by the lights at <math>C</math> and <math>D</math> are equal. To an observer at <math>C</math>, the angle determined by the lights at <math>A</math> and <math>B</math> and the angle determined by the lights at <math>D</math> and <math>B</math> are equal. The number of kilometers from <math>A</math> to <math>D</math> is given by <math>\frac {p\sqrt{q}}{r}</math>, where <math>p</math>, <math>q</math>, and <math>r</math> are relatively prime positive integers, and <math>r</math> is not divisible by the square of any prime. Find <math>p</math> + <math>q</math> + <math>r</math>. | ||
+ | ==Diagram== | ||
+ | <asy> | ||
+ | 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)); | ||
+ | </asy> | ||
+ | -asjpz | ||
− | == Solution == | + | ==Video Solution by Punxsutawney Phil== |
+ | https://www.youtube.com/watch?v=f3zEesJh4Ws | ||
− | Let O be the intersection of BC and AD. By the Angle Bisector Theorem,5/ | + | == Solution 1== |
+ | |||
+ | Let <math>O</math> be the intersection of <math>BC</math> and <math>AD</math>. By the [[Angle Bisector Theorem]], <math>\frac {5}{BO}</math> = <math>\frac {13}{CO}</math>, so <math>BO</math> = <math>5x</math> and <math>CO</math> = <math>13x</math>, and <math>BO</math> + <math>OC</math> = <math>BC</math> = <math>12</math>, so <math>x</math> = <math>\frac {2}{3}</math>, and <math>OC</math> = <math>\frac {26}{3}</math>. Let <math>P</math> be the foot of the altitude from <math>D</math> to <math>OC</math>. It can be seen that triangle <math>DOP</math> is similar to triangle <math>AOB</math>, and triangle <math>DPC</math> is similar to triangle <math>ABC</math>. If <math>DP</math> = <math>15y</math>, then <math>CP</math> = <math>36y</math>, <math>OP</math> = <math>10y</math>, and <math>OD</math> = <math>5y\sqrt {13}</math>. Since <math>OP</math> + <math>CP</math> = <math>46y</math> = <math>\frac {26}{3}</math>, <math>y</math> = <math>\frac {13}{69}</math>, and <math>AD</math> = <math>\frac {60\sqrt{13}}{23}</math> (by the pythagorean theorem on triangle <math>ABO</math> we sum <math>AO</math> and <math>OD</math>). The answer is <math>60</math> + <math>13</math> + <math>23</math> = <math>\boxed{096}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Extend <math>AB</math> and <math>CD</math> to intersect at <math>P</math>. Note that since <math>\angle ACB=\angle PCB</math> and <math>\angle ABC=\angle PBC=90^{\circ}</math> by ASA congruency we have <math>\triangle ABC\cong \triangle PBC</math>. Therefore <math>AC=PC=13</math>. | ||
+ | |||
+ | By the angle bisector theorem, <math>PD=\dfrac{130}{23}</math> and <math>CD=\dfrac{169}{23}</math>. Now we apply Stewart's theorem to find <math>AD</math>: | ||
+ | |||
+ | <cmath>\begin{align*}13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=13\cdot 13\cdot \dfrac{130}{23}+10\cdot 10\cdot \dfrac{169}{23}\\ | ||
+ | 13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=\dfrac{169\cdot 130+169\cdot 100}{23}\\ | ||
+ | 13\cdot \dfrac{130}{23}\cdot \dfrac{169}{23}+13\cdot AD^2&=1690\\ | ||
+ | AD^2&=130-\dfrac{130\cdot 169}{23^2}\\ | ||
+ | AD^2&=\dfrac{130\cdot 23^2-130\cdot 169}{23^2}\\ | ||
+ | AD^2&=\dfrac{130(23^2-169)}{23^2}\\ | ||
+ | AD^2&=\dfrac{130(360)}{23^2}\\ | ||
+ | AD&=\dfrac{60\sqrt{13}}{23}\\ | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | and our final answer is <math>60+13+23=\boxed{096}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | Notice that by extending <math>AB</math> and <math>CD</math> to meet at a point <math>E</math>, <math>\triangle ACE</math> is isosceles. Now we can do a straightforward coordinate bash. Let <math>C=(0,0)</math>, <math>B=(12,0)</math>, <math>E=(12,-5)</math> and <math>A=(12,5)</math>, and the equation of line <math>CD</math> is <math>y=-\dfrac{5}{12}x</math>. Let F be the intersection point of <math>AD</math> and <math>BC</math>, and by using the Angle Bisector Theorem: <math>\dfrac{BF}{AB}=\dfrac{FC}{AC}</math> we have <math>FC=\dfrac{26}{3}</math>. Then the equation of the line <math>AF</math> through the points <math>(12,5)</math> and <math>\left(\frac{26}{3},0\right)</math> is <math>y=\frac32 x-13</math>. Hence the intersection point of <math>AF</math> and <math>CD</math> is the point <math>D</math> at the coordinates <math>\left(\dfrac{156}{23},-\dfrac{65}{23}\right)</math>. Using the distance formula, <math>AD=\sqrt{\left(12-\dfrac{156}{23}\right)^2+\left(5+\dfrac{65}{23}\right)^2}=\dfrac{60\sqrt{13}}{23}</math> for an answer of <math>60+13+23=\fbox{096}</math>. | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | After drawing a good diagram, we reflect <math>ABC</math> over the line <math>BC</math>, forming a new point that we'll call <math>A'</math>. Also, let the intersection of <math>AD</math> and <math>BC</math> be point <math>E</math>. Point <math>D</math> lies on line <math>A'C</math>. Since line <math>AD</math> bisects <math>\angle{CAB}</math>, we can use the Angle Bisector Theorem. <math>AA'=10</math> and <math>AC=13</math>, so <math>\frac{CD}{A'D}=\frac{13}{10}</math>. Letting the segments be <math>13x</math> and <math>10x</math> respectively, we now have <math>13x+10x=13</math>. Therefore, <math>x=\frac{13}{23}</math>. By the Pythagorean Theorem, <math>AE=\frac{5\sqrt{13}}{3}</math>. Using the Angle Bisector Theorem on <math>\angle{ACD}</math>, we have that <math>ED=\frac{5x\sqrt{13}}{3}</math>. Substituting in <math>x=\frac{13}{23}</math>, we have that <math>AD=\left(\frac{5\sqrt{13}}{3}\right)(1+x)=\frac{60\sqrt{13}}{23}</math>, so the answer is <math>60+13+23=\boxed{096}</math>. | ||
+ | |||
+ | -RootThreeOverTwo | ||
+ | |||
+ | ==Solution 5 (Angle Bisector Theorem + Law of Cosines)== | ||
+ | Let <math> CD</math> and <math> AB</math> meet at <math> A'</math>; then <math> \overline{AD}</math> is an angle bisector of isosceles <math> \triangle AA'C</math>. Then by the Angle Bisector Theorem, <math> A'D=\frac {10}{10+13} \cdot 13=\frac {130}{23}</math>, and <math> \cos \angle DA'A=\frac {5}{13}</math>. By the Law of Cosines on <math> \triangle AA'D</math>, we have <cmath>AD^2=10^2+\left(\frac {130}{23}\right)^2-2 \cdot 10 \cdot \frac {130}{23} \cdot \frac {5}{13}=\frac {2^4 \cdot 3^2 \cdot 5^2 \cdot 13}{23^2} \Longrightarrow AD=\frac {60\sqrt {13}}{23}</cmath> and the answer is <math> 60+13+23=\boxed{096}</math>. | ||
+ | |||
+ | |||
+ | <asy> | ||
+ | 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)); | ||
+ | </asy> | ||
+ | |||
+ | -azjps | ||
+ | |||
+ | ==Solution 6 (Law of Sines)== | ||
+ | <asy>pathpen = linewidth(0.7); | ||
+ | pen f = fontsize(10); | ||
+ | size(144); | ||
+ | pair B=(0,0); | ||
+ | pair A=(5,0); | ||
+ | pair C=(0,12); | ||
+ | pair E=(-5,0); | ||
+ | pair abA = A+unit(C-A)+unit(B-A); | ||
+ | pair D = extension(A,abA,C,E); | ||
+ | D(A--C); | ||
+ | D(C--B); | ||
+ | D(A--B); | ||
+ | D(C--D); | ||
+ | D(A--D); | ||
+ | D(D--E,dashed); | ||
+ | D(B--E,dashed); | ||
+ | D(rightanglemark(C,B,A,20),red); | ||
+ | MP("A",D(A),plain.SE,f); | ||
+ | MP("B",D(B),plain.S,f); | ||
+ | MP("C",D(C),plain.N,f); | ||
+ | MP("D",D(D),plain.NW,f); | ||
+ | MP("E",D(E),plain.SW,f); | ||
+ | MP("5",(A+B)/2,plain.S,f); | ||
+ | MP("12",(B+C)/2,plain.E,f); | ||
+ | MP("13",(A+C)/2,plain.NE,f); | ||
+ | MP("\alpha",A,4*(unit(D-A)+unit(C-A))/2,f); | ||
+ | MP("\alpha",A,4*(unit(B-A)+unit(D-A))/2,f); | ||
+ | MP("2\alpha",E,4*(unit(C-E)+unit(B-E))/2,f); | ||
+ | MP("\beta",C,6*(unit(A-C)+unit(B-C))/2,f); | ||
+ | MP("\beta",C,6*(unit(E-C)+unit(B-C))/2,f);</asy> | ||
+ | |||
+ | Using the law of sines on <math> \bigtriangleup ADE</math>, | ||
+ | |||
+ | <cmath>\begin{align*} | ||
+ | AD &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(180^\circ-3\alpha)}\\ | ||
+ | &= AE \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha+2\alpha)}\\ | ||
+ | &= 10 \cdot \dfrac{\sin(2\alpha)}{\sin(\alpha)\cos(2\alpha)+\cos(\alpha)\sin(2\alpha)}\\ | ||
+ | &= 10 \cdot \dfrac{\sin(2\alpha)}{\sqrt{\dfrac{1-\cos(2\alpha)}{2}}\cdot\cos(2\alpha)+\sqrt{\dfrac{1+\cos(2\alpha)}{2}}\cdot\sin(2\alpha)}\\ | ||
+ | &= 10 \cdot \dfrac{\dfrac{12}{13}}{\sqrt{\dfrac{8}{26}}\cdot\dfrac{5}{13}+\sqrt{\dfrac{18}{26}}\cdot\dfrac{12}{13}}\\ | ||
+ | &= 10 \cdot \dfrac{12\sqrt{13}}{10+36} = \dfrac{60\sqrt{13}}{23} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | <math> \therefore</math> the answer is <math> 60+13+23=\boxed{096}</math>. | ||
+ | |||
+ | -m1sterzer0 | ||
+ | |||
+ | == See Also == | ||
+ | |||
+ | {{AIME box|year=2009|n=II|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 13:56, 18 December 2022
Contents
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
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=f3zEesJh4Ws
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
.
-RootThreeOverTwo
Solution 5 (Angle Bisector Theorem + Law of Cosines)
Let and
meet at
; then
is an angle bisector of isosceles
. Then by the Angle Bisector Theorem,
, and
. By the Law of Cosines on
, we have
and the answer is
.
-azjps
Solution 6 (Law of Sines)
Using the law of sines on ,
the answer is
.
-m1sterzer0
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.