Difference between revisions of "2010 AIME I Problems/Problem 15"
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Blueprimes (talk | contribs) (→Solution 6 (Similar to Solution 1 with easier computation)) |
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== Problem == | == Problem == | ||
− | In <math>\triangle{ABC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such that the [[incircle]]s of <math>\triangle{ABM}</math> and <math>\triangle{BCM}</math> have equal [[inradius|radii]]. | + | In <math>\triangle{ABC}</math> with <math>AB = 12</math>, <math>BC = 13</math>, and <math>AC = 15</math>, let <math>M</math> be a point on <math>\overline{AC}</math> such that the [[incircle]]s of <math>\triangle{ABM}</math> and <math>\triangle{BCM}</math> have equal [[inradius|radii]]. Then <math>\frac{AM}{CM} = \frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p + q</math>. |
== Solution == | == Solution == | ||
Line 30: | Line 30: | ||
label("$12$",(2.28834,5.75684),NE); | label("$12$",(2.28834,5.75684),NE); | ||
dot((7.33333,0)); | dot((7.33333,0)); | ||
− | label("$M$",(7.56053,- | + | label("$M$",(7.56053,-1.000),NE); |
+ | label("$H_1$",(3.97942,-1.200),NE); | ||
+ | label("$H_2$",(9.54741,-1.200),NE); | ||
dot((4.66667,2.49444)); | dot((4.66667,2.49444)); | ||
label("$I_1$",(3.97942,2.92179),NE); | label("$I_1$",(3.97942,2.92179),NE); | ||
dot((9.66667,2.49444)); | dot((9.66667,2.49444)); | ||
− | label("$I_2$",( | + | label("$I_2$",(9.54741,2.92179),NE); |
clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle); | clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle); | ||
</asy></center> | </asy></center> | ||
Line 42: | Line 44: | ||
By [[Stewart's Theorem]], we have <math>12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)</math> or <math>432 = 3d^2 + 40x - 3x^2.</math> Brute forcing by plugging in our previous result for <math>d</math>, we have <math>432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.</math> Clearing the fraction and gathering like terms, we get <math>0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.</math> | By [[Stewart's Theorem]], we have <math>12^2(15 - x) + 13^2x = d^215 + 15x(15 - x)</math> or <math>432 = 3d^2 + 40x - 3x^2.</math> Brute forcing by plugging in our previous result for <math>d</math>, we have <math>432 = \frac {3(25x - 180)^2}{(15 - 2x)^2} + 40x - 3x^2.</math> Clearing the fraction and gathering like terms, we get <math>0 = 12x^4 - 340x^3 + 2928x^2 - 7920x.</math> | ||
− | ''Aside: Since <math>x</math> must be rational in order for our answer to be in the desired form, we can use the [[Rational Root Theorem]] to reveal that <math> | + | ''Aside: Since <math>x</math> must be rational in order for our answer to be in the desired form, we can use the [[Rational Root Theorem]] to reveal that <math>6x</math> is an integer because we can divide the polynomial by <math>2</math>. The only such <math>x</math> in the above-stated range is <math>\frac {22}3</math>.'' |
Legitimately solving that quartic, note that <math>x = 0</math> and <math>x = 15</math> should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get <math>0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).</math> The only solution in the desired range is thus <math>\frac {22}3</math>. Then <math>CM = \frac {23}3</math>, and our desired ratio <math>\frac {AM}{CM} = \frac {22}{23}</math>, giving us an answer of <math>\boxed{045}</math>. | Legitimately solving that quartic, note that <math>x = 0</math> and <math>x = 15</math> should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get <math>0 = 4x(x - 15)(3x^2 - 40x + 132) = x(x - 15)(x - 6)(3x - 22).</math> The only solution in the desired range is thus <math>\frac {22}3</math>. Then <math>CM = \frac {23}3</math>, and our desired ratio <math>\frac {AM}{CM} = \frac {22}{23}</math>, giving us an answer of <math>\boxed{045}</math>. | ||
Line 113: | Line 115: | ||
So <math>AM=22/3</math> or <math>6</math>. But from (1) we know that <math>5+15(AM-7)>0</math>, or <math>AM>7-1/3>6</math>, so <math>AM=22/3</math>, <math>CM=15-22/3=23/3</math>, <math>AM/CM=22/23</math>. | So <math>AM=22/3</math> or <math>6</math>. But from (1) we know that <math>5+15(AM-7)>0</math>, or <math>AM>7-1/3>6</math>, so <math>AM=22/3</math>, <math>CM=15-22/3=23/3</math>, <math>AM/CM=22/23</math>. | ||
− | '' | + | === Solution 5 === |
+ | |||
+ | Let the common inradius equal r, <math>BM = x</math>, <math>AM = y</math>, <math>MC = z</math> | ||
+ | |||
+ | From the prespective of <math>\triangle{ABM}</math> and <math>\triangle{BMC}</math> we get: | ||
+ | |||
+ | <math>S_{ABM} = rs = r \cdot (\frac{12+x+y}{2})</math> <math>S_{BMC} = rs = r \cdot (\frac{13+x+z}{2})</math> | ||
+ | |||
+ | Add two triangles up, we get <math>\triangle{ABC}</math> : | ||
+ | |||
+ | <math>S_{ABC} = S_{ABM} + S_{BMC} = r \cdot \frac{25+2x+y+z}{2}</math> | ||
+ | |||
+ | Since <math>y + z = 15</math>, we get: | ||
+ | |||
+ | <math>r = \frac{S_{ABC}}{20 + x}</math> | ||
+ | |||
+ | By drawing an altitude from <math>I_1</math> down to a point <math>H_1</math> and from <math>I_2</math> to <math>H_2</math>, we can get: | ||
+ | |||
+ | <math>r \cdot cot(\frac{\angle A}{2}) =r \cdot A H_1 = r \cdot \frac{12+y-x}{2} </math> and | ||
+ | |||
+ | <math>r \cdot cot(\frac{\angle C}{2}) = r \cdot H_2 C = r \cdot \frac{13+z-x}{2}</math> | ||
+ | |||
+ | Adding these up, we get: | ||
+ | |||
+ | <math>r \cdot (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) = \frac{25+y+z-2x}{2} = \frac{40-2x}{2} = 20-x</math> | ||
+ | |||
+ | <math>r = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}</math> | ||
+ | |||
+ | Now, we have 2 values equal to r, we can set them equal to each other: | ||
+ | |||
+ | <math>\frac{S_{ABC}}{20 + x} = \frac{20-x}{cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})}</math> | ||
+ | |||
+ | If we let R denote the incircle of ABC, note: | ||
+ | |||
+ | AC = <math>(cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R = 15</math> and | ||
+ | |||
+ | <math>S_{ABC} = \frac{12+13+15}{2} \cdot R = 20 \cdot R</math>. | ||
+ | |||
+ | By cross multiplying the equation above, we get: | ||
+ | |||
+ | <math>400 - x^2 = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot S_{ABC} = (cot(\frac{\angle A}{2})+cot(\frac{\angle C}{2})) \cdot R \cdot 20 = 15 \cdot 20 = 300</math> | ||
+ | |||
+ | We can find out x: | ||
+ | |||
+ | <math>x = 10</math>. | ||
+ | |||
+ | Now, we can find ratio of y and z: | ||
+ | |||
+ | <math>\frac{AM}{CM} = \frac{y}{z} = \frac{S_{ABM}}{S_{BCM}} = \frac{r \cdot \frac{22+y}{2} }{r \cdot \frac{23+z}{2}} = \frac{22+y}{23+z} = \frac{22}{23}</math> | ||
+ | |||
+ | The answer is <math>\boxed{045}</math>. | ||
+ | |||
+ | -Alexlikemath | ||
+ | |||
+ | === Solution 6 (Similar to Solution 1 with easier computation) === | ||
+ | |||
+ | Let <math>CM=x, AM=rx, BM=d</math>. <math>x+rx=15\Rightarrow x=\frac{15}{1+r}</math>. | ||
+ | |||
+ | Similar to Solution 1, we have | ||
+ | <cmath> | ||
+ | r=\frac{[AMB]}{[CMB]}=\frac{12+rx+d}{13+x+d} \Rightarrow d=\frac{13r-12}{1-r} | ||
+ | </cmath> | ||
+ | as well as | ||
+ | <cmath> | ||
+ | 12^2\cdot x + 13^2 rx=15x\cdot rx+15d^2 (\text{via Stewart's Theorem}) | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | \frac{(12^2 + 13^2r) \cdot 15}{1+r} - \frac{15r\cdot 15^2}{(1+r)^2}=\frac{15(13r-12)^2}{(1-r)^2} | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | \frac{169r^2+88r+144}{(1+r)^2}=\frac{(13r-12)^2}{(1-r)^2} | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | (169r^2+88r+144)((r^2+1)-2r)=(169r^2-312r+144)((r^2+1)+2r) | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | (r^2+1)(400r)=2r(338r^2-224r+288) | ||
+ | </cmath> | ||
+ | <cmath> | ||
+ | 100(r^2+1)=169r^2-112r+144 \Rightarrow 69r^2-112r+44=(23r-22)(3r-2)=0 | ||
+ | </cmath> | ||
+ | |||
+ | Since <math>d=\frac{13r-12}{1-r}>0</math>, we have <math>r=\frac{22}{23} \longrightarrow \boxed{045}</math>. | ||
+ | |||
+ | ~ asops | ||
+ | |||
+ | === Solution 7 (No Stewart's or trig, fast + clever) === | ||
+ | |||
+ | Let <math>BM = d, AM = x, CM = 15 - x</math>. Observe that we have the equation by the incircle formula: | ||
+ | <cmath>\frac{[ABM]}{12 + AM + MB} = \frac{[CBM]}{13 + CM + MB} \implies \frac{AM}{CM} = \frac{12 + MB}{13 + MB} \implies \frac{x}{15 - x} = \frac{12 + d}{13 + d}.</cmath> | ||
+ | Now let <math>X</math> be the point of tangency between the incircle of <math>\triangle ABC</math> and <math>AC</math>. Additionally, let <math>P</math> and <math>Q</math> be the points of tangency between the incircles of <math>\triangle ABM</math> and <math>\triangle CBM</math> with <math>AC</math> respectively. Some easy calculation yields <math>AX = 7, CX = 8</math>. By homothety we have | ||
+ | <cmath>\frac{AP}{7} = \frac{CQ}{8} \implies 8(AP) = 7(CQ) \implies 8(12 + x - d) = 7(13 + 15 - x - d) \implies d = 15x - 100.</cmath> | ||
+ | Substituting into the first equation derived earlier it is left to solve | ||
+ | <cmath>\frac{x}{15 - x} = \frac{15x - 88}{15x - 87} \implies 3x^2 - 40x + 132 \implies (x - 6)(3x - 22) = 0.</cmath> | ||
+ | Now <math>x = 6</math> yields <math>d = -10</math> which is invalid, hence <math>x = \frac{22}{3}</math> so <math>\frac{AM}{CM} = \frac{\frac{22}{3}}{15 - \frac{22}{3}} = \frac{22}{23}.</math> The requested sum is <math>22 + 23 = \boxed{45}</math>. ~blueprimes | ||
− | |||
==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=UQVI0Q2PWZw&feature=youtu.be&fbclid=IwAR338IdppnZVuze4rzT0gm8G2NB_uIsmj175WgD6sa43gg3EhFAGm5bAvV0 | https://www.youtube.com/watch?v=UQVI0Q2PWZw&feature=youtu.be&fbclid=IwAR338IdppnZVuze4rzT0gm8G2NB_uIsmj175WgD6sa43gg3EhFAGm5bAvV0 | ||
+ | |||
+ | == Sidenote == | ||
+ | In the problem, <math>BM=10</math> and the equal inradius of the two triangles happens to be <math> \frac {2\sqrt{14}}{3}</math>. | ||
== See Also == | == See Also == |
Latest revision as of 13:14, 11 January 2024
Contents
Problem
In with
,
, and
, let
be a point on
such that the incircles of
and
have equal radii. Then
, where
and
are relatively prime positive integers. Find
.
Solution
![[asy] /* from geogebra: see azjps, userscripts.org/scripts/show/72997 */ import graph; defaultpen(linewidth(0.7)+fontsize(10)); size(200); /* segments and figures */ draw((0,0)--(15,0)); draw((15,0)--(6.66667,9.97775)); draw((6.66667,9.97775)--(0,0)); draw((7.33333,0)--(6.66667,9.97775)); draw(circle((4.66667,2.49444),2.49444)); draw(circle((9.66667,2.49444),2.49444)); draw((4.66667,0)--(4.66667,2.49444)); draw((9.66667,2.49444)--(9.66667,0)); /* points and labels */ label("r",(10.19662,1.92704),SE); label("r",(5.02391,1.8773),SE); dot((0,0)); label("$A$",(-1.04408,-0.60958),NE); dot((15,0)); label("$C$",(15.41907,-0.46037),NE); dot((6.66667,9.97775)); label("$B$",(6.66525,10.23322),NE); label("$15$",(6.01866,-1.15669),NE); label("$13$",(11.44006,5.50815),NE); label("$12$",(2.28834,5.75684),NE); dot((7.33333,0)); label("$M$",(7.56053,-1.000),NE); label("$H_1$",(3.97942,-1.200),NE); label("$H_2$",(9.54741,-1.200),NE); dot((4.66667,2.49444)); label("$I_1$",(3.97942,2.92179),NE); dot((9.66667,2.49444)); label("$I_2$",(9.54741,2.92179),NE); clip((-3.72991,-6.47862)--(-3.72991,17.44518)--(32.23039,17.44518)--(32.23039,-6.47862)--cycle); [/asy]](http://latex.artofproblemsolving.com/6/5/0/65068f7716651b40ecb18b0763a47e58b9bc3434.png)
Solution 1
Let , then
. Also let
Clearly,
. We can also express each area by the rs formula. Then
. Equating and cross-multiplying yields
or
Note that for
to be positive, we must have
.
By Stewart's Theorem, we have or
Brute forcing by plugging in our previous result for
, we have
Clearing the fraction and gathering like terms, we get
Aside: Since must be rational in order for our answer to be in the desired form, we can use the Rational Root Theorem to reveal that
is an integer because we can divide the polynomial by
. The only such
in the above-stated range is
.
Legitimately solving that quartic, note that and
should clearly be solutions, corresponding to the sides of the triangle and thus degenerate cevians. Factoring those out, we get
The only solution in the desired range is thus
. Then
, and our desired ratio
, giving us an answer of
.
Solution 2
Let and
so
. Let the incenters of
and
be
and
respectively, and their equal inradii be
. From
, we find that
Let the incircle of meet
at
and the incircle of
meet
at
. Then note that
is a rectangle. Also,
is right because
and
are the angle bisectors of
and
respectively and
. By properties of tangents to circles
and
. Now notice that the altitude of
to
is of length
, so by similar triangles we find that
(3). Equating (3) with (1) and (2) separately yields
and adding these we have
Solution 3
Let the incircle of hit
,
,
at
, and let the incircle of
hit
,
,
at
. Draw the incircle of
, and let it be tangent to
at
. Observe that we have a homothety centered at A sending the incircle of
to that of
, and one centered at
taking the incircle of
to that of
. These have the same power, since they take incircles of the same size to the same circle. Also, the power of the homothety is
.
By standard computations, and
. Now, let
and
. We will now go around and chase lengths. Observe that
. Then,
. We also have
, so
and
.
Observe now that . Also,
. Solving, we get
and
(as a side note, note that
, a result that I actually believe appears in Mandelbrot 1995-2003, or some book in that time-range).
Now, we get . To finish, we will compute area ratios.
. Also, since their inradii are equal, we get
. Equating and cross multiplying yields the quadratic
, so
. However, observe that
, so we take
. Our ratio is therefore
, giving the answer
.
Note: Once we have and
, it's bit easier to use use the right triangle of
than chasing the area ratio. The inradius of
can be calculated to be
, and the inradius of
and
are
, so,
or,
We get
or
.
Solution 4
Suppose the incircle of touches
at
, and the incircle of
touches
at
. Then
We have ,
,
,
Therefore
And since ,
,
Now,
So or
. But from (1) we know that
, or
, so
,
,
.
Solution 5
Let the common inradius equal r, ,
,
From the prespective of and
we get:
Add two triangles up, we get :
Since , we get:
By drawing an altitude from down to a point
and from
to
, we can get:
and
Adding these up, we get:
Now, we have 2 values equal to r, we can set them equal to each other:
If we let R denote the incircle of ABC, note:
AC = and
.
By cross multiplying the equation above, we get:
We can find out x:
.
Now, we can find ratio of y and z:
The answer is .
-Alexlikemath
Solution 6 (Similar to Solution 1 with easier computation)
Let .
.
Similar to Solution 1, we have
as well as
Since , we have
.
~ asops
Solution 7 (No Stewart's or trig, fast + clever)
Let . Observe that we have the equation by the incircle formula:
Now let
be the point of tangency between the incircle of
and
. Additionally, let
and
be the points of tangency between the incircles of
and
with
respectively. Some easy calculation yields
. By homothety we have
Substituting into the first equation derived earlier it is left to solve
Now
yields
which is invalid, hence
so
The requested sum is
. ~blueprimes
Video Solution
Sidenote
In the problem, and the equal inradius of the two triangles happens to be
.
See Also
2010 AIME I (Problems • Answer Key • Resources) | ||
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