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Difference between revisions of "2001 USAMO Problems/Problem 2"
m (→‎Solution 3: filled in a missing parenthesis :))
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Thus, <math>Q'</math> has normalized <math>x</math>-coordinate <math>1-\frac{s-a}{s}=\frac{a}{s}</math>.
 
Thus, <math>Q'</math> has normalized <math>x</math>-coordinate <math>1-\frac{s-a}{s}=\frac{a}{s}</math>.
  
−
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>
+
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>
  
 
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>.
 
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>.

Revision as of 21:24, 14 March 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 $AB+BD_3=AB + BF=AF = AG = AC + AG=AC + CD_3$. 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.

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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