Difference between revisions of "2019 AIME II Problems/Problem 15"
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Y = IP(c, P--(2*Q-P)); | Y = IP(c, P--(2*Q-P)); | ||
draw(A--B--C--A, black+0.8); | draw(A--B--C--A, black+0.8); | ||
− | draw(c^^X--Y^^B-- | + | draw(c^^X--Y^^B--Q^^C--P); |
dot("$A$", A, SW); | dot("$A$", A, SW); | ||
dot("$B$", B, SE); | dot("$B$", B, SE); | ||
Line 29: | Line 29: | ||
dot("$Y$", Y, NW); | dot("$Y$", Y, NW); | ||
label("$25$", P--Q, SW); | label("$25$", P--Q, SW); | ||
− | label("$15$", | + | label("$15$", Q--Y, SW); |
− | label("$10$", | + | label("$10$", X--P, SW); |
</asy> | </asy> | ||
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First we have <math>a\cos A=PQ=25</math>, and <math>(a\cos A)(c\cos C)=(a\cos C)(c\cos A)=AP\cdot PB=10(25+15)=400</math> by PoP. Similarly, <math>(a\cos A)(b\cos B)=15(10+25)=525,</math> and dividing these each by <math>a\cos A</math> gives | First we have <math>a\cos A=PQ=25</math>, and <math>(a\cos A)(c\cos C)=(a\cos C)(c\cos A)=AP\cdot PB=10(25+15)=400</math> by PoP. Similarly, <math>(a\cos A)(b\cos B)=15(10+25)=525,</math> and dividing these each by <math>a\cos A</math> gives | ||
<math>b\cos B=21,c\cos C=16</math>. | <math>b\cos B=21,c\cos C=16</math>. | ||
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It is known that the sides of the orthic triangle are <math>a\cos A,b\cos B,c\cos C</math>, and its angles are <math>\pi-2A</math>,<math>\pi-2B</math>, and <math>\pi-2C</math>. We thus have the three sides of the orthic triangle now. | It is known that the sides of the orthic triangle are <math>a\cos A,b\cos B,c\cos C</math>, and its angles are <math>\pi-2A</math>,<math>\pi-2B</math>, and <math>\pi-2C</math>. We thus have the three sides of the orthic triangle now. | ||
Letting <math>D</math> be the foot of the altitude from <math>A</math>, we have, in <math>\triangle DPQ</math>, | Letting <math>D</math> be the foot of the altitude from <math>A</math>, we have, in <math>\triangle DPQ</math>, | ||
− | <cmath>\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}=27 | + | <cmath>\cos P,\cos Q=\frac{21^2+25^2-16^2}{2\cdot 21\cdot 25},\frac{16^2+25^2-21^2}{2\cdot 16\cdot 25}= \frac{27}{35}, \frac{11}{20}.</cmath> |
− | <cmath>\Rightarrow \cos B=\cos\ | + | <cmath>\Rightarrow \cos B=\cos\left(\tfrac 12 (\pi-P)\right)=\sin\tfrac 12 P =\sqrt{\frac{4}{35}},</cmath> |
similarly, we get | similarly, we get | ||
− | <cmath>\cos C=\cos\ | + | <cmath>\cos C=\cos\left(\tfrac 12 (\pi-Q)\right)=\sin\tfrac 12 Q=\sqrt{\frac{9}{40}}.</cmath> |
− | To finish, <cmath>bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40}) | + | To finish, <cmath>bc= \frac{(b\cos B)(c\cos C)}{\cos B\cos C}=\frac{16\cdot 21}{(2/\sqrt{35})(3/\sqrt{40})}=560\sqrt{14}.</cmath> |
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The requested sum is <math>\boxed{574}</math>. | The requested sum is <math>\boxed{574}</math>. | ||
+ | - crazyeyemoody907 | ||
− | + | Remark: The proof that <math>a \cos A = PQ</math> can be found here: http://www.irmo.ie/5.Orthic_triangle.pdf | |
==Solution 2== | ==Solution 2== | ||
− | Let <math> | + | Let <math>BC=a</math>, <math>AC=b</math>, and <math>AB=c</math>. Let <math>\cos\angle A=k</math>. Then <math>AP=bk</math> and <math>AQ=ck</math>. |
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− | So the final answer is <math>560 + 14 = \boxed{574} </math> | + | By Power of a Point theorem, |
+ | <cmath>\begin{align} | ||
+ | AP\cdot BP=XP\cdot YP \quad &\Longrightarrow \quad b^2k^2-bck+400=0\\ | ||
+ | AQ\cdot CQ=YQ\cdot XQ \quad &\Longrightarrow \quad c^2k^2-bck+525=0 | ||
+ | \end{align}</cmath> | ||
+ | Thus <math>bck = (bk)^2+400=(ck)^2+525 = u</math>. Then <math>bk=\sqrt{u-400}</math>, <math>ck=\sqrt{u-525}</math>, and | ||
+ | <cmath>k=\sqrt{\frac{(u-400)(u-525)}{u^2}}</cmath> | ||
+ | Use the Law of Cosines in <math>\triangle APQ</math> to get <math>25^2=b^2k^2+c^2k^2-2bck^3 = 2bck-925-2bck^3</math>, which rearranges to <cmath>775=bck - k^2\cdot bck = u-\frac{(u-400)(u-525)}{u}</cmath>Upon simplification, this reduces to a linear equation in <math>u</math>, with solution <math>u=1400</math>. Then <cmath>AB\cdot AC = bc = \frac 1{k}\cdot bck = \frac{u^2}{\sqrt{(u-400)(u-525)}}=560 \sqrt{14}</cmath> | ||
+ | So the final answer is <math>560 + 14 = \boxed{574}</math> | ||
By SpecialBeing2017 | By SpecialBeing2017 | ||
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==Solution 3== | ==Solution 3== | ||
− | Let <math>\ | + | Let <math>AP=p</math>, <math>PB=q</math>, <math>AQ=r</math>, and <math>QC=s</math>. By Power of a Point, |
+ | <cmath>\begin{align} | ||
+ | AP\cdot PB=XP\cdot YP \quad &\Longrightarrow \quad pq=400\\ | ||
+ | AQ\cdot QC=YQ\cdot XQ \quad &\Longrightarrow \quad rs=525 | ||
+ | \end{align}</cmath> | ||
+ | Points <math>P</math> and <math>Q</math> lie on the circle, <math>\omega</math>, with diameter <math>BC</math>, and pow<math>(A,\omega) = AP\cdot AB = AQ\cdot AC</math>, so <cmath> p(p+q)=r(r+s)\quad \Longrightarrow \quad p^2-r^2=125</cmath> Use Law of Cosines in <math>\triangle APQ</math> to get <math>25^2=p^2+r^2-2pr\cos A</math>; since <math>\cos A = \frac r{p+q}</math>, this simplifies as | ||
+ | <cmath>500 \ =\ 2r^2-\frac{2pr^2}{p+q} \ =\ 2r^2-\frac{2p^2r^2}{p^2+400} \ =\ \frac{800r^2}{r^2+525}</cmath> | ||
+ | We get <math>r=5\sqrt{35}</math> and thus | ||
+ | <cmath>r=5\sqrt{35}, \quad p = \sqrt{r^2+125} = 10\sqrt{10}, \quad q = \frac{400}{p} =4\sqrt{10}, \quad s= \frac{525}{r} = 3\sqrt{35}.</cmath> | ||
+ | Therefore <math>AB\cdot AC = (p+q)\cdot(r+s) = 560\sqrt{14}</math>. So the answer is <math>560 + 14 = \boxed{574}</math> | ||
− | By | + | By asr41 |
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− | + | ==Solution 4 (Clean)== | |
+ | This solution is directly based of @CantonMathGuy's solution. | ||
+ | We start off with a key claim. | ||
− | <math> | + | <i> Claim. </i> <math>XB \parallel AC</math> and <math>YC \parallel AB</math>. |
− | < | + | <i> Proof. </i> |
+ | [[File:AIME-II-2019-15.png|350px|right]] | ||
− | + | Let <math>E</math> and <math>F</math> denote the reflections of the orthocenter over points <math>P</math> and <math>Q</math>, respectively. Since <math>EF \parallel XY</math> and <cmath>EF = 2 PQ = XP + PQ + QY = XY,</cmath> we have that <math>E X Y F</math> is a rectangle. Then, since <math>\angle XYF = 90^\circ</math> we obtain <math>\angle XBF = 90^\circ</math> (which directly follows from <math>XBYF</math> being cyclic); hence <math>\angle XBQ = \angle AQB</math>, or <math>XB \parallel AQ \Rightarrow XB \parallel AC</math>. | |
− | + | Similarly, we can obtain <math>YC \parallel AB</math>. <math>\ \blacksquare</math> | |
− | + | A direct result of this claim is that <math>\triangle BPX \sim \triangle APQ \sim \triangle CYQ</math>. | |
− | + | Thus, we can set <math>AP = 5k</math> and <math>BP = 2k</math>, then applying Power of a Point on <math>P</math> we get <math>10 \cdot 40 = 10k^2 \implies k = 2\sqrt{10} \implies AB = 14 \sqrt{10}</math>. Also, we can set <math>AQ = 5l</math> and <math>CQ = 3l</math> and once again applying Power of a Point (but this time to <math>Q</math>) we get | |
− | <math> | ||
− | + | <math>\phantom{...................}15 \cdot 35 = 15l^2 \implies l = \sqrt{35} \implies AC = 8 \sqrt{35}</math>. | |
− | + | Hence, | |
− | + | <math>\phantom{...................}AB \cdot AC = 112 \sqrt{350} = 112 \cdot 5 \sqrt{14} = 560 \sqrt{14}</math> | |
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− | + | and the answer is <math>560 + 14 = \boxed{574}</math>. ~rocketsri | |
+ | ==Solution 5== | ||
+ | [[File:2019AIMEIIP15Solution.png|900px]] | ||
+ | '''mathboy282''' | ||
− | + | ==Video Solution by MOP 2024== | |
+ | https://youtu.be/aYV09qIwTqs | ||
+ | ~r00tsOfUnity | ||
==See Also== | ==See Also== |
Latest revision as of 03:11, 7 January 2024
Contents
Problem
In acute triangle points
and
are the feet of the perpendiculars from
to
and from
to
, respectively. Line
intersects the circumcircle of
in two distinct points,
and
. Suppose
,
, and
. The value of
can be written in the form
where
and
are positive integers, and
is not divisible by the square of any prime. Find
.
Diagram
Solution 1
First we have , and
by PoP. Similarly,
and dividing these each by
gives
.
It is known that the sides of the orthic triangle are , and its angles are
,
, and
. We thus have the three sides of the orthic triangle now.
Letting
be the foot of the altitude from
, we have, in
,
similarly, we get
To finish,
The requested sum is
.
- crazyeyemoody907
Remark: The proof that can be found here: http://www.irmo.ie/5.Orthic_triangle.pdf
Solution 2
Let ,
, and
. Let
. Then
and
.
By Power of a Point theorem,
Thus
. Then
,
, and
Use the Law of Cosines in
to get
, which rearranges to
Upon simplification, this reduces to a linear equation in
, with solution
. Then
So the final answer is
By SpecialBeing2017
Solution 3
Let ,
,
, and
. By Power of a Point,
Points
and
lie on the circle,
, with diameter
, and pow
, so
Use Law of Cosines in
to get
; since
, this simplifies as
We get
and thus
Therefore
. So the answer is
By asr41
Solution 4 (Clean)
This solution is directly based of @CantonMathGuy's solution. We start off with a key claim.
Claim. and
.
Proof.
Let and
denote the reflections of the orthocenter over points
and
, respectively. Since
and
we have that
is a rectangle. Then, since
we obtain
(which directly follows from
being cyclic); hence
, or
.
Similarly, we can obtain .
A direct result of this claim is that .
Thus, we can set and
, then applying Power of a Point on
we get
. Also, we can set
and
and once again applying Power of a Point (but this time to
) we get
.
Hence,
and the answer is . ~rocketsri
Solution 5
Video Solution by MOP 2024
~r00tsOfUnity
See Also
2019 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last Question | |
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.