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

Revision as of 05:09, 22 September 2015

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 $AQ = D_2P$ .

we use barycentric coordinates. It’s obvious that un-normalized, D1 = (0 : s − c : s − b) ⇒ D2 = (0 : s − b : s − c), so we get a normalized D2 =

0, s−b a , s−c a � . Similarly, E2 =

s−a b , 0, s−c b � . Now we obtain the points P =

s−a s , s−b s , s−c s � by intersecting the lines AD2 : (s−c)y = (s−b)z and BE2 : (s − c)x = (s − a)z. Let Q0 be such that AQ0 = P D2. It’s obvious that Q0 y + Py = Ay + D2y, so we find that Q0 y = s−b a − s−b s = (s−a)(s−b) sa . Also, since it lies on the line AD2, we get that Q0 z = s−c s−b · Q0 y = (s−a)(s−c) sa . Hence, Q 0 x = 1 − ((s − b) + (s − c))(s − a) sa = sa − a(s − a) sa = a s Hence, Q 0 = � a s , (s − a)(s − b) sa , (s − a)(s − c) sa � 16 A B C I D1 E1 D2 E2 Q P Figure 5.1: USAMO 2001/2 Let I =

a 2s , b 2s , c 2s � . We claim that, in fact, I is the midpoint of Q0D1. Indeed, 1 2 � 0 + a s � = a 2s 1 2 � (s − a)(s − b) sa + s − c a � = (s − a)(s − b) + s(s − c) 2sa = ab 2sa = b 2s 1 2 � (s − a)(s − c) sa + s − b a � = c 2s Implying that Q lies on the circle; in particular, diametrically opposite from D1, so it is the closer of the two points. Hence, Q = Q0 , so we’re done.

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

Revision as of 05:09, 22 September 2015

Problem

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@@ Line 2: / Line 2: @@
 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>.
-== Solution ==
-=== 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>.)
-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>.
+we use barycentric coordinates.
+It’s obvious that un-normalized, D1 = (0 : s − c : s − b) ⇒ D2 = (0 : s − b : s − c), so we get a
+normalized D2 =
-<center><asy>
+,
-pathpen = linewidth(0.7); size(300); pen d = linetype("4 4") + linewidth(0.6);
+s−b
-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 */
+a
+,
+s−c
+a
+�
+. Similarly, E2 =
-/* ugly construction for OA */
+s−a
-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);
+b
-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));
+, 0,
+s−c
+b
+�
+.
+Now we obtain the points P =
-clip((-20,-10)--(-20,20)--(20,20)--(20,-10)--cycle);
+s−a
-</asy></center>
+s
+,
+s−b
+s
+,
+s−c
+s
+�
+by intersecting the lines AD2 : (s−c)y = (s−b)z
+and BE2 : (s − c)x = (s − a)z.
+Let Q0 be such that AQ0 = P D2. It’s obvious that Q0
+y + Py = Ay + D2y, so we find that Q0
+y =
+s−b
+a −
+s−b
+s =
+(s−a)(s−b)
+sa
+. Also, since it lies on the line AD2, we get that Q0
+z =
+s−c
+s−b
+· Q0
+y =
+(s−a)(s−c)
+sa
+.
+Hence,
+Q
+x = 1 −
+((s − b) + (s − c))(s − a)
+sa
+=
+sa − a(s − a)
+sa
+=
+a
+s
+Hence,
+Q
+=
+�
+a
+s
+,
+(s − a)(s − b)
+sa
+,
+(s − a)(s − c)
+sa �
+A
+B C
+I
+D1
+E1
+D2
+E2
+Q
+P
+Figure 5.1: USAMO 2001/2
+Let I =
-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>
+a
+s
-=== Solution 2===
+,
-The key observation is the following lemma.
+b
+s
-'''Lemma''': Segment <math>D_1Q</math> is a diameter of circle <math>\omega</math>.
+,
-<center>[[File:2001usamo2-1.png]]</center>
+c
-''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>.
+s
+�
-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>.
+. We claim that, in fact, I is the midpoint of Q0D1. Indeed,
-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>
+a
-It follows that
+s
-<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>
+a
-that is, <math>D' = D_2</math>, as desired. <math>\blacksquare</math>
+s
-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>.
+(s − a)(s − b)
+sa
-<center>[[File:2001usamo2-2.png]]</center>
++
+s − c
-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
+a
-<cmath>\frac{IM_1}{AP} = \frac{GM_1}{AG} = \frac{1}{2}</cmath>
+�
-or
+=
-<cmath>AP = 2IM_1.</cmath>
+(s − a)(s − b) + s(s − c)
-This yields
+sa
-<cmath>AQ = AP - QP = 2IM_1 - QP = QD_2 - QP = PD_2,</cmath>
+=
-as desired.
+ab
+sa
-'''Note''': We used directed lengths in our calculations to avoid possible complications caused by the different shapes of triangle <math>ABC</math>.
+=
+b
+s
+�
+(s − a)(s − c)
+sa
++
+s − b
+a
+�
+=
+c
+s
+Implying that Q lies on the circle; in particular, diametrically opposite from D1, so it is the
+closer of the two points. Hence, Q = Q0
+, so we’re done.
 == See also ==