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

Latest revision as of 07:52, 22 September 2025

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 $\frac{CE_2}{E_2A} \cdot \frac{AP}{PD_2} \cdot \frac{BD_2}{BC} = 1 \Longrightarrow \frac{AP}{PD_2} = \frac{(b - (s-a)) \cdot a}{(a-(s-b)) \cdot AE_1} = \frac{a}{s-a}$ . 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.

@@ Line 10: / Line 10: @@
 <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>
+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);
-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>
+</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{(b - (s-a)) \cdot a}{(a-(s-b)) \cdot AE_1} = \frac{a}{s-a}</math>. It easily follows that <math>AQ = D_2P</math>. <math>\blacksquare</math>
 ===Solution 2===

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