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Difference between revisions of "2001 USAMO Problems/Problem 2"
(→‎Solution: official solution)
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== Solution ==
 
== 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 + AG=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>.)
 
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 + AG=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>.)
  
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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>
 
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>
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=== Solution 2===
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The key observation is the following lemma.
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 +
'''Lemma''': Segment <math>D_1Q</math> is a diameter of circle <math>\omega</math>.
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{{image}}
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''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>.
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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>.
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It is well known that
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<cmath>CD_1 = \frac{1}{2}(BC + CA - AB).</cmath>
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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
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<cmath>AX = AY = \frac{1}{2}(AX + AY) = \frac{1}{2}(AB + BX + YC + CA) = \frac{1}{2}(AB + BC + CA).</cmath>
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It follows that
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<cmath>BD' = BX = AX - AB = \frac{1}{2}(BC + CA - AB).</cmath>
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Combining these two equations yields <math>BD' = CD_1</math>. Thus
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<cmath>BD_2 = BD_1 - D_2D_1 = D_2C - D_2D_1 = CD_1 = BD',</cmath>
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that is, <math>D' = D_2</math>, as desired. <math>\blacksquare</math>
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{{image}}
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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
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<cmath>QD_2 = 2IM_1</cmath>
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and <math>AD_2\parallel M_1I</math>. Similarly, we can prove that <math>BE_2\parallel M_2I</math>.
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{{image}}
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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
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<cmath>\frac{IM_1}{AP} = \frac{GM_1}{AG} = \frac{1}{2}</cmath>
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or
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<cmath>AP = 2IM_1.</cmath>
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This yields
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<cmath>AQ = AP - QP = 2IM_1 - QP = QD_2 - QP = PD_2,</cmath>
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as desired.
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'''Note''': We used directed lengths in our calculations to avoid possible complications caused by the different shapes of triangle <math>ABC</math>.
  
 
== See also ==
 
== See also ==

Revision as of 20:54, 6 July 2014

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 $\omega$.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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. $\blacksquare$

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $BE_2\parallel M_2I$.

An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.

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 $ABC$.

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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