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Difference between revisions of "2001 USAMO Problems/Problem 2"

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Let <math>ABC</math> be a [[triangle]] and let <math>\omega</math> be its [[incircle]]. Denote by <math>D_1</math> and <math>E_1</math> the points where <math>\omega</math> is tangent to sides <math>BC</math> and <math>AC</math>, respectively. Denote by <math>D_2</math> and <math>E_2</math> the points on sides <math>BC</math> and <math>AC</math>, respectively, such that <math>CD_2 = BD_1</math> and <math>CE_2 = AE_1</math>, and denote by <math>P</math> the point of intersection of segments <math>AD_2</math> and <math>BE_2</math>. Circle <math>\omega</math> intersects segment <math>AD_2</math> at two points, the closer of which to the vertex <math>A</math> is denoted by <math>Q</math>. Prove that <math>AQ = D_2P</math>.
 
Let <math>ABC</math> be a [[triangle]] and let <math>\omega</math> be its [[incircle]]. Denote by <math>D_1</math> and <math>E_1</math> the points where <math>\omega</math> is tangent to sides <math>BC</math> and <math>AC</math>, respectively. Denote by <math>D_2</math> and <math>E_2</math> the points on sides <math>BC</math> and <math>AC</math>, respectively, such that <math>CD_2 = BD_1</math> and <math>CE_2 = AE_1</math>, and denote by <math>P</math> the point of intersection of segments <math>AD_2</math> and <math>BE_2</math>. Circle <math>\omega</math> intersects segment <math>AD_2</math> at two points, the closer of which to the vertex <math>A</math> is denoted by <math>Q</math>. Prove that <math>AQ = D_2P</math>.
  
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== Solution ==
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==Solution==
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{{solution}}
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===Solution 1===
 +
 
 +
It is well known that the excircle opposite <math>A</math> is tangent to <math>\overline{BC}</math> at the point <math>D_2</math>. (Proof: let the points of tangency of the excircle with the lines <math>BC, AB, AC</math> be <math>D_3, F,G</math> respectively. Then <math>AB+BD_3=AB + BF=AF = AG = AC + CG=AC + CD_3</math>. It follows that <math>2CD_3 = AB + BC - AC</math>, and <math>CD_3 = s-b = BD_1 = CD_2</math>, so <math>D_3 \equiv D_2</math>.)
 +
 
 +
Now consider the homothety that carries the incircle of <math>\triangle ABC</math> to its excircle. The homothety also carries <math>Q</math> to <math>D_2</math> (since <math>A,Q,D_2</math> are collinear), and carries the tangency points <math>E_1</math> to <math>G</math>. It follows that <math>\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}</math>.
 +
 
 +
<asy>
 +
pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP("D_1",foot(O,B,C))),E1 = D(MP("E_1",foot(O,A,C),NE)),E2 = D(MP("E_2",C+A-E1,NE)); /* arbitrary points */  /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca,C,B), OA = IP(A--A+10*(O-A),C--C+50*(Cb-C)), D2 = D(MP("D_2",foot(OA,B,C))), Fa=2B-A, Ga=2C-A, F=MP("F",D(foot(OA,B,Fa)),NW), G=MP("G",D(foot(OA,C,Ga)),NE); D(OA); D(MP("A",A,N)--MP("B",B,NW)--MP("C",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D(B--Fa,linewidth(0.6)); D(C--Ga,linewidth(0.6)); D(MP("P",IP(D(A--D2),D(B--E2)),NNE)); D(MP("Q",IP(incircle(A,B,C),A--D2),SW));  clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle); </asy>
 +
By Menelaus' Theorem on <math>\triangle ACD_2</math> with segment <math>\overline{BE_2}</math>, it follows that <math>\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(c - (s-a)) \cdot a}{(a-(s-c)) \cdot AE_1} = \frac{a}{s-a}</math>. It easily follows that <math>AQ = D_2P</math>. <math>\blacksquare</math>
 +
 
 +
===Solution 2===
 +
 
 +
The key observation is the following lemma.
 +
 
 +
Lemma: Segment <math>D_1Q</math> is a diameter of circle <math>\omega</math>.
 +
 
 +
Proof: Let <math>I</math> be the center of circle <math>\omega</math>, i.e., <math>I</math> is the incenter of triangle <math>ABC</math>. Extend segment <math>D_1I</math> through <math>I</math> to intersect circle <math>\omega</math> again at <math>Q'</math>, and extend segment <math>AQ'</math> through <math>Q'</math> to intersect segment <math>BC</math> at <math>D'</math>. We show that <math>D_2 = D'</math>, which in turn implies that <math>Q = Q'</math>, that is, <math>D_1Q</math> is a diameter of <math>\omega</math>.
 +
 
 +
Let <math>l</math> be the line tangent to circle <math>\omega</math> at <math>Q'</math>, and let <math>l</math> intersect the segments <math>AB</math> and <math>AC</math> at <math>B_1</math> and <math>C_1</math>, respectively. Then <math>\omega</math> is an excircle of triangle <math>AB_1C_1</math>. Let <math>\mathbf{H}_1</math> denote the dilation with its center at <math>A</math> and ratio <math>AD'/AQ'</math>. Since <math>l\perp D_1Q'</math> and <math>BC\perp D_1Q'</math>, <math>l\parallel BC</math>. Hence <math>AB/AB_1 = AC/AC_1 = AD'/AQ'</math>. Thus <math>\mathbf{H}_1(Q') = D'</math>, <math>\mathbf{H}_1(B_1) = B</math>, and <math>\mathbf{H}_1(C_1) = C</math>. It also follows that an excircle <math>\Omega</math> of triangle <math>ABC</math> is tangent to the side <math>BC</math> at <math>D'</math>.
 +
 
 +
It is well known that <cmath>CD_1 = \frac{1}{2}(BC + CA - AB).</cmath> We compute <math>BD'</math>. Let <math>X</math> and <math>Y</math> denote the points of tangency of circle <math>\Omega</math> with rays <math>AB</math> and <math>AC</math>, respectively. Then by equal tangents, <math>AX = AY</math>, <math>BD' = BX</math>, and <math>D'C = YC</math>. Hence <cmath>AX = AY = \frac{1}{2}(AX + AY) = \frac{1}{2}(AB + BX + YC + CA) = \frac{1}{2}(AB + BC + CA).</cmath> It follows that <cmath>BD' = BX = AX - AB = \frac{1}{2}(BC + CA - AB).</cmath> Combining these two equations yields <math>BD' = CD_1</math>. Thus <cmath>BD_2 = BD_1 - D_2D_1 = D_2C - D_2D_1 = CD_1 = BD',</cmath> that is, <math>D' = D_2</math>, as desired. <math>\blacksquare</math>
 +
 
 +
Now we prove our main result. Let <math>M_1</math> and <math>M_2</math> be the respective midpoints of segments <math>BC</math> and <math>CA</math>. Then <math>M_1</math> is also the midpoint of segment <math>D_1D_2</math>, from which it follows that <math>IM_1</math> is the midline of triangle <math>D_1QD_2</math>. Hence <cmath>QD_2 = 2IM_1</cmath> and <math>AD_2\parallel M_1I</math>. Similarly, we can prove that <math>BE_2\parallel M_2I</math>.
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 +
2001usamo2-2.png
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Let <math>G</math> be the centroid of triangle <math>ABC</math>. Thus segments <math>AM_1</math> and <math>BM_2</math> intersect at <math>G</math>. Define transformation <math>\mathbf{H}_2</math> as the dilation with its center at <math>G</math> and ratio <math>-1/2</math>. Then <math>\mathbf{H}_2(A) = M_1</math> and <math>\mathbf{H}_2(B) = M_2</math>. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since <math>AD_2\parallel M_1I</math> and <math>BE_2\parallel M_2I</math>, <math>\mathbf{H}_2</math> maps lines <math>AD_2</math> and <math>BE_2</math> to lines <math>M_1I</math> and <math>M_2I</math>, respectively. It also follows that <math>\mathbf{H}_2(I) = P</math> and <cmath>\frac{IM_1}{AP} = \frac{GM_1}{AG} = \frac{1}{2}</cmath> or <cmath>AP = 2IM_1.</cmath> This yields <cmath>AQ = AP - QP = 2IM_1 - QP = QD_2 - QP = PD_2,</cmath> as desired.
 +
 
 +
Note: We used directed lengths in our calculations to avoid possible complications caused by the different shapes of triangle <math>ABC</math>.
 +
 
 +
=== Solution 3 ===
 +
 
 +
Here is a rather nice solution using barycentric coordinates:
 +
 
 +
Let <math>A</math> be <math>(1,0,0)</math>, <math>B</math> be <math>(0,1,0)</math>, and <math>C</math> be <math>(0,0,1)</math>. Let the side lengths of the triangle be <math>a,b,c</math> and the semi-perimeter <math>s</math>.
 +
 
 +
Now, <cmath>CD_1=s-c, BD_1=s-b, AE_1=s-a, CE_1=s-c.</cmath> Thus, <cmath>CD_2=s-b, BD_2=s-c, AE_2=s-c, CE_2=s-a.</cmath>
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Therefore, <math>D_2=(0:s-b:s-c)</math> and <math>E_2=(s-a:0:s-c).</math> Clearly then, the non-normalized coordinates of <math>P=(s-a:s-b:s-c).</math>
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 +
Normalizing, we have that <cmath>D_2=\left(0,\frac{s-b}{a},\frac{s-c}{a}\right), E_2=\left(\frac{s-a}{b},0,\frac{s-c}{b}\right), P=\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right).</cmath>
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Now, we find the point <math>Q'</math> inside the triangle on the line <math>AD_2</math> such that <math>AQ'=D_2P</math>. It is then sufficient to show that this point lies on the incircle.
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<math>P</math> is the fraction <math>\frac{s-a}{s}</math> of the way "up" the line segment from <math>D_2</math> to <math>A</math>. Thus, we are looking for the point that is <math>\frac{s-a}{s}</math> of the way "down" the line segment from <math>A</math> to <math>D_2</math>, or, the fraction <math>1-\frac{s-a}{s}</math> of the way "up".
 +
 
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Thus, <math>Q'</math> has normalized <math>x</math>-coordinate <math>1-\frac{s-a}{s}=\frac{a}{s}</math>.
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 +
As the line <math>AD_2</math> has equation <math>(s-c)y=(s-b)z</math>, it can easily be found that <cmath>Q'=\left(\frac{a^2}{as}, \frac{(s-a)(s-b)}{as}, \frac{(s-a)(s-c)}{as}\right)=(a^2:(s-a)(s-b):(s-a)(s-c)).</cmath>
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Recalling that the equation of the incircle is <cmath>a^2yz+b^2xz+c^2xy+(x+y+z)[(s-a)^2x+(s-b)^2y+(s-c)^2z]=0.</cmath> We must show that this equation is true for <math>Q'</math>'s values of <math>x,y,z</math>.
 +
 
 +
Plugging in our values, this means showing that
 +
<cmath>a^2(s-a)^2(s-b)(s-c)+a^2(s-a)[b^2(s-c)+c^2(s-b)]+as[a^2(s-a)^2+(s-a)(s-b)^3+(s-a)(s-c)^3]=0.</cmath>
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Dividing by <math>a(s-a)</math>, this is just
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<cmath>a(s-a)(s-b)(s-c)+a[b^2(s-c)+c^2(s-b)]+a^2s(s-a)+s(s-b)^3+s(s-c)^3=0.</cmath>
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Plugging in the value of <math>s:</math>
 +
<cmath>\frac{a(-a+b+c)(a-b+c)(a+b-c)}{8}+\frac{ab^2(a+b-c)}{2}+\frac{ac^2(a-b+c)}{2}+\frac{a^2(a+b+c)(-a+b+c)}{4}+\frac{(a+b+c)(a-b+c)^3}{16}+\frac{(a+b+c)(a+b-c)^3}{16}=0.</cmath>
 +
<cmath>2a(-a+b+c)(a-b+c)(a+b-c)+8ab^2(a+b-c)+8ac^2(a-b+c)+4a^2(a+b+c)(-a+b+c)+(a+b+c)(a-b+c)^3+(a+b+c)(a+b-c)^3=0</cmath>
 +
<cmath>2a[(-a+b+c)(a-b+c)(a+b-c)+4b^2(a+b-c)+4c^2(a-b+c)]+(a+b+c)[4a^2(-a+b+c)+(a-b+c)^3+(a+b-c)^3]=0</cmath>
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The first bracket is just
 +
<cmath>-a^3-b^3-c^3+a^2b+a^2c+b^2c+b^2a+c^2a+c^2b-2abc+4ab^2+4b^3-4b^2c+4ac^2-4bc^2+4c^3</cmath>
 +
<cmath>=-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc</cmath>
 +
and the second bracket is <cmath>-2a^3+4a^2b+4a^2c+6ab^2+6ac^2-12abc.</cmath>
 +
Dividing everything by <math>2a</math> gives
 +
<cmath>-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc+(a+b+c)(-a^2+2ab+2ac+3b^2+3c^2-6bc),</cmath> which is <math>0</math>, as desired.
 +
 
 +
As <math>Q'</math> lies on the incircle and <math>AD_2</math>, <math>Q'=Q</math>, and our proof is complete.
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 +
 
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=== Solution 4 ===
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We again use Barycentric coordinates. As before, let <math>A</math> be <math>(1,0,0)</math>, <math>B</math> be <math>(0,1,0)</math>, and <math>C</math> be <math>(0,0,1)</math>. Also
 +
 
 +
<cmath>D_{1}=\left( 0,\frac{s-c}{a},\frac{s-b}{a} \right), D_2=\left(0,\frac{s-b}{a},\frac{s-c}{a}\right), E_2=\left(\frac{s-a}{b},0,\frac{s-c}{b}\right), P=\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right).</cmath>
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Now, consider a point <math>Q'</math> for which <math>AQ'=PD_{2}</math>. Then working component-vise, we get <math>Q'+P=A+D_{2}</math> from which we can easily get the coordinates of <math>Q'</math> as;
 +
 
 +
<cmath>Q'=\left(\frac{a}{s},\frac{(s-a)(s-b)}{sa},\frac{(s-a)(s-c)}{sa}\right)</cmath>
 +
 
 +
It suffices to show that <math>Q'=Q</math>.
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 +
Let <math>I=\left(\frac{a}{2s},\frac{b}{2s},\frac{c}{2s}\right)</math>  be the incenter of triangle <math>ABC</math>. We claim that <math>I</math> is the midpoint of <math>Q'D_{1}</math>. Indeed,
 +
 
 +
<cmath>\frac{1}{2}\left(0+\frac{a}{s}\right)=\frac{a}{2s}</cmath>
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 +
<cmath>\frac{1}{2}\left(\frac{(s-a)(s-b)}{sa}+\frac{s-c}{a}\right)=\frac{(s-a)(s-b)+s(s-c)}{2sa}=\frac{ab}{2sa}=\frac{b}{2s}</cmath>
 +
 
 +
<cmath>\frac{1}{2}\left(\frac{(s-a)(s-c)}{sa}+\frac{s-b}{a}\right)=\frac{c}{2s}</cmath>
 +
 
 +
Hence the claim has been proved.
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 +
Since <math>I</math> is the center of <math>\omega</math> and the midpoint of  <math>Q'D_{1}</math>, thus <math>Q'</math> is the point diametrically opposite to <math>D_{1}</math>, and hence it lies on <math>\omega</math> and closer to <math>A</math> off the 2 points. Thus <math>Q=Q'</math> as desired.
  
 
== See also ==
 
== See also ==
Line 9: Line 100:
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 07:35, 16 November 2017

Problem

Let $ABC$ be a triangle and let $\omega$ be its incircle. Denote by $D_1$ and $E_1$ the points where $\omega$ is tangent to sides $BC$ and $AC$, respectively. Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$, respectively, such that $CD_2 = BD_1$ and $CE_2 = AE_1$, and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$. Circle $\omega$ intersects segment $AD_2$ at two points, the closer of which to the vertex $A$ is denoted by $Q$. Prove that $AQ = D_2P$.

Solution

Solution 1

It is well known that the excircle opposite $A$ is tangent to $\overline{BC}$ at the point $D_2$. (Proof: let the points of tangency of the excircle with the lines $BC, AB, AC$ be $D_3, F,G$ respectively. Then $AB+BD_3=AB + BF=AF = AG = AC + CG=AC + CD_3$. It follows that $2CD_3 = AB + BC - AC$, and $CD_3 = s-b = BD_1 = CD_2$, so $D_3 \equiv D_2$.)

Now consider the homothety that carries the incircle of $\triangle ABC$ to its excircle. The homothety also carries $Q$ to $D_2$ (since $A,Q,D_2$ are collinear), and carries the tangency points $E_1$ to $G$. It follows that $\frac{AQ}{QD_2} = \frac{AE_1}{E_1G} = \frac{s-a}{E_1C + CD_2} = \frac{s-a}{CD_1 + BD_1} = \frac{s-a}{a}$.

[asy] pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6); pair B=(0,0), C=(10,0), A=7*expi(1),O=D(incenter(A,B,C)),D1 = D(MP("D_1",foot(O,B,C))),E1 = D(MP("E_1",foot(O,A,C),NE)),E2 = D(MP("E_2",C+A-E1,NE)); /* arbitrary points */ /* ugly construction for OA */ pair Ca = 2C-A, Cb = bisectorpoint(Ca,C,B), OA = IP(A--A+10*(O-A),C--C+50*(Cb-C)), D2 = D(MP("D_2",foot(OA,B,C))), Fa=2B-A, Ga=2C-A, F=MP("F",D(foot(OA,B,Fa)),NW), G=MP("G",D(foot(OA,C,Ga)),NE); D(OA); D(MP("A",A,N)--MP("B",B,NW)--MP("C",C,NE)--cycle); D(incircle(A,B,C)); D(CP(OA,D2),d); D(B--Fa,linewidth(0.6)); D(C--Ga,linewidth(0.6)); D(MP("P",IP(D(A--D2),D(B--E2)),NNE)); D(MP("Q",IP(incircle(A,B,C),A--D2),SW)); clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle); [/asy] By Menelaus' Theorem on $\triangle ACD_2$ with segment $\overline{BE_2}$, it follows that $\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(c - (s-a)) \cdot a}{(a-(s-c)) \cdot AE_1} = \frac{a}{s-a}$. It easily follows that $AQ = D_2P$. $\blacksquare$

Solution 2

The key observation is the following lemma.

Lemma: Segment $D_1Q$ is a diameter of circle $\omega$.

Proof: Let $I$ be the center of circle $\omega$, i.e., $I$ is the incenter of triangle $ABC$. Extend segment $D_1I$ through $I$ to intersect circle $\omega$ again at $Q'$, and extend segment $AQ'$ through $Q'$ to intersect segment $BC$ at $D'$. We show that $D_2 = D'$, which in turn implies that $Q = Q'$, that is, $D_1Q$ is a diameter of $\omega$.

Let $l$ be the line tangent to circle $\omega$ at $Q'$, and let $l$ intersect the segments $AB$ and $AC$ at $B_1$ and $C_1$, respectively. Then $\omega$ is an excircle of triangle $AB_1C_1$. Let $\mathbf{H}_1$ denote the dilation with its center at $A$ and ratio $AD'/AQ'$. Since $l\perp D_1Q'$ and $BC\perp D_1Q'$, $l\parallel BC$. Hence $AB/AB_1 = AC/AC_1 = AD'/AQ'$. Thus $\mathbf{H}_1(Q') = D'$, $\mathbf{H}_1(B_1) = B$, and $\mathbf{H}_1(C_1) = C$. It also follows that an excircle $\Omega$ of triangle $ABC$ is tangent to the side $BC$ at $D'$.

It is well known that \[CD_1 = \frac{1}{2}(BC + CA - AB).\] We compute $BD'$. Let $X$ and $Y$ denote the points of tangency of circle $\Omega$ with rays $AB$ and $AC$, respectively. Then by equal tangents, $AX = AY$, $BD' = BX$, and $D'C = YC$. Hence \[AX = AY = \frac{1}{2}(AX + AY) = \frac{1}{2}(AB + BX + YC + CA) = \frac{1}{2}(AB + BC + CA).\] It follows that \[BD' = BX = AX - AB = \frac{1}{2}(BC + CA - AB).\] Combining these two equations yields $BD' = CD_1$. Thus \[BD_2 = BD_1 - D_2D_1 = D_2C - D_2D_1 = CD_1 = BD',\] that is, $D' = D_2$, as desired. $\blacksquare$

Now we prove our main result. Let $M_1$ and $M_2$ be the respective midpoints of segments $BC$ and $CA$. Then $M_1$ is also the midpoint of segment $D_1D_2$, from which it follows that $IM_1$ is the midline of triangle $D_1QD_2$. Hence \[QD_2 = 2IM_1\] and $AD_2\parallel M_1I$. Similarly, we can prove that $BE_2\parallel M_2I$.

2001usamo2-2.png Let $G$ be the centroid of triangle $ABC$. Thus segments $AM_1$ and $BM_2$ intersect at $G$. Define transformation $\mathbf{H}_2$ as the dilation with its center at $G$ and ratio $-1/2$. Then $\mathbf{H}_2(A) = M_1$ and $\mathbf{H}_2(B) = M_2$. Under the dilation, parallel lines go to parallel lines and the intersection of two lines goes to the intersection of their images. Since $AD_2\parallel M_1I$ and $BE_2\parallel M_2I$, $\mathbf{H}_2$ maps lines $AD_2$ and $BE_2$ to lines $M_1I$ and $M_2I$, respectively. It also follows that $\mathbf{H}_2(I) = P$ and \[\frac{IM_1}{AP} = \frac{GM_1}{AG} = \frac{1}{2}\] or \[AP = 2IM_1.\] This yields \[AQ = AP - QP = 2IM_1 - QP = QD_2 - QP = PD_2,\] as desired.

Note: We used directed lengths in our calculations to avoid possible complications caused by the different shapes of triangle $ABC$.

Solution 3

Here is a rather nice solution using barycentric coordinates:

Let $A$ be $(1,0,0)$, $B$ be $(0,1,0)$, and $C$ be $(0,0,1)$. Let the side lengths of the triangle be $a,b,c$ and the semi-perimeter $s$.

Now, \[CD_1=s-c, BD_1=s-b, AE_1=s-a, CE_1=s-c.\] Thus, \[CD_2=s-b, BD_2=s-c, AE_2=s-c, CE_2=s-a.\]

Therefore, $D_2=(0:s-b:s-c)$ and $E_2=(s-a:0:s-c).$ Clearly then, the non-normalized coordinates of $P=(s-a:s-b:s-c).$

Normalizing, we have that \[D_2=\left(0,\frac{s-b}{a},\frac{s-c}{a}\right), E_2=\left(\frac{s-a}{b},0,\frac{s-c}{b}\right), P=\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right).\]

Now, we find the point $Q'$ inside the triangle on the line $AD_2$ such that $AQ'=D_2P$. It is then sufficient to show that this point lies on the incircle.

$P$ is the fraction $\frac{s-a}{s}$ of the way "up" the line segment from $D_2$ to $A$. Thus, we are looking for the point that is $\frac{s-a}{s}$ of the way "down" the line segment from $A$ to $D_2$, or, the fraction $1-\frac{s-a}{s}$ of the way "up".

Thus, $Q'$ has normalized $x$-coordinate $1-\frac{s-a}{s}=\frac{a}{s}$.

As the line $AD_2$ has equation $(s-c)y=(s-b)z$, it can easily be found that \[Q'=\left(\frac{a^2}{as}, \frac{(s-a)(s-b)}{as}, \frac{(s-a)(s-c)}{as}\right)=(a^2:(s-a)(s-b):(s-a)(s-c)).\]

Recalling that the equation of the incircle is \[a^2yz+b^2xz+c^2xy+(x+y+z)[(s-a)^2x+(s-b)^2y+(s-c)^2z]=0.\] We must show that this equation is true for $Q'$'s values of $x,y,z$.

Plugging in our values, this means showing that \[a^2(s-a)^2(s-b)(s-c)+a^2(s-a)[b^2(s-c)+c^2(s-b)]+as[a^2(s-a)^2+(s-a)(s-b)^3+(s-a)(s-c)^3]=0.\] Dividing by $a(s-a)$, this is just \[a(s-a)(s-b)(s-c)+a[b^2(s-c)+c^2(s-b)]+a^2s(s-a)+s(s-b)^3+s(s-c)^3=0.\]

Plugging in the value of $s:$ \[\frac{a(-a+b+c)(a-b+c)(a+b-c)}{8}+\frac{ab^2(a+b-c)}{2}+\frac{ac^2(a-b+c)}{2}+\frac{a^2(a+b+c)(-a+b+c)}{4}+\frac{(a+b+c)(a-b+c)^3}{16}+\frac{(a+b+c)(a+b-c)^3}{16}=0.\] \[2a(-a+b+c)(a-b+c)(a+b-c)+8ab^2(a+b-c)+8ac^2(a-b+c)+4a^2(a+b+c)(-a+b+c)+(a+b+c)(a-b+c)^3+(a+b+c)(a+b-c)^3=0\] \[2a[(-a+b+c)(a-b+c)(a+b-c)+4b^2(a+b-c)+4c^2(a-b+c)]+(a+b+c)[4a^2(-a+b+c)+(a-b+c)^3+(a+b-c)^3]=0\] The first bracket is just \[-a^3-b^3-c^3+a^2b+a^2c+b^2c+b^2a+c^2a+c^2b-2abc+4ab^2+4b^3-4b^2c+4ac^2-4bc^2+4c^3\] \[=-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc\] and the second bracket is \[-2a^3+4a^2b+4a^2c+6ab^2+6ac^2-12abc.\] Dividing everything by $2a$ gives \[-a^3+3b^3+3c^3+a^2b+a^2c-3b^2c+5b^2a+5c^2a-3c^2b-2abc+(a+b+c)(-a^2+2ab+2ac+3b^2+3c^2-6bc),\] which is $0$, as desired.

As $Q'$ lies on the incircle and $AD_2$, $Q'=Q$, and our proof is complete.


Solution 4

We again use Barycentric coordinates. As before, let $A$ be $(1,0,0)$, $B$ be $(0,1,0)$, and $C$ be $(0,0,1)$. Also

\[D_{1}=\left( 0,\frac{s-c}{a},\frac{s-b}{a} \right), D_2=\left(0,\frac{s-b}{a},\frac{s-c}{a}\right), E_2=\left(\frac{s-a}{b},0,\frac{s-c}{b}\right), P=\left(\frac{s-a}{s},\frac{s-b}{s},\frac{s-c}{s}\right).\]

Now, consider a point $Q'$ for which $AQ'=PD_{2}$. Then working component-vise, we get $Q'+P=A+D_{2}$ from which we can easily get the coordinates of $Q'$ as;

\[Q'=\left(\frac{a}{s},\frac{(s-a)(s-b)}{sa},\frac{(s-a)(s-c)}{sa}\right)\]

It suffices to show that $Q'=Q$.

Let $I=\left(\frac{a}{2s},\frac{b}{2s},\frac{c}{2s}\right)$ be the incenter of triangle $ABC$. We claim that $I$ is the midpoint of $Q'D_{1}$. Indeed,

\[\frac{1}{2}\left(0+\frac{a}{s}\right)=\frac{a}{2s}\]

\[\frac{1}{2}\left(\frac{(s-a)(s-b)}{sa}+\frac{s-c}{a}\right)=\frac{(s-a)(s-b)+s(s-c)}{2sa}=\frac{ab}{2sa}=\frac{b}{2s}\]

\[\frac{1}{2}\left(\frac{(s-a)(s-c)}{sa}+\frac{s-b}{a}\right)=\frac{c}{2s}\]

Hence the claim has been proved.

Since $I$ is the center of $\omega$ and the midpoint of $Q'D_{1}$, thus $Q'$ is the point diametrically opposite to $D_{1}$, and hence it lies on $\omega$ and closer to $A$ off the 2 points. Thus $Q=Q'$ as desired.

See also

2001 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

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