Difference between revisions of "2014 USAJMO Problems/Problem 6"

Latest revision as of 17:17, 28 August 2021

Problem

Let $ABC$ be a triangle with incenter $I$ , incircle $\gamma$ and circumcircle $\Gamma$ . Let $M,N,P$ be the midpoints of sides $\overline{BC}$ , $\overline{CA}$ , $\overline{AB}$ and let $E,F$ be the tangency points of $\gamma$ with $\overline{CA}$ and $\overline{AB}$ , respectively. Let $U,V$ be the intersections of line $EF$ with line $MN$ and line $MP$ , respectively, and let $X$ be the midpoint of arc $BAC$ of $\Gamma$ .

(a) Prove that $I$ lies on ray $CV$ .

(b) Prove that line $XI$ bisects $\overline{UV}$ .

Solution

$[asy] unitsize(5cm); import olympiad; pair A, B, C, I, M, N, P, E, F, U, V, X, R; A = dir(190); B = dir(120); C = dir(350); I = incenter(A, B, C); label("$A$", A, W); label("$B$", B, dir(90)); label("$C$", C, dir(0)); dot(I); label("$I$", I, SSE); draw(A--B--C--cycle); real r, R; r = inradius(A, B, C); R = circumradius(A, B, C); path G, g; G = circumcircle(A, B, C); g = incircle(A, B, C); draw(G); draw(g); label("$\Gamma$", dir(35), dir(35)); label("$\gamma$", 2/3 * dir(125)); M = (B+C)/2; N = (A+C)/2; P = (A+B)/2; label("$M$", M, NE); label("$N$", N, SE); label("$P$", P, W); E = tangent(A, I, r, 1); F = tangent(A, I, r, 2); label("$E$", E, SW); label("$F$", F, WNW); U = extension(E, F, M, N); V = intersectionpoint(P--M, F--E); label("$U$", U, S); label("$V$", V, NE); draw(P--M--U--F); X = dir(235); label("$X$", X, dir(235)); draw(X--I, dashed); draw(C--V, dashed); draw(A--I); label("$x^\circ$", A + (0.2,0), dir(90)); label("$y^\circ$", C + (-0.4,0), dir(90)); [/asy]$

(a)

Solution 1: We will prove this via contradiction: assume that line $IC$ intersects line $MP$ at $Q$ and line $EF$ and $R$ , with $R$ and $Q$ not equal to $V$ . Let $x = \angle A/2 = \angle IAE$ and $y = \angle C/2 = \angle ICA$ . We know that $\overline{MP} \parallel \overline{AC}$ because $MP$ is a midsegment of triangle $ABC$ ; thus, by alternate interior angles (A.I.A) $\angle MVE = \angle FEA = (180^\circ - 2x) / 2 = 90^\circ - x$ , because triangle $AFE$ is isosceles. Also by A.I.A, $\angle MQC = \angle QCA = y$ . Furthermore, because $AI$ is an angle bisector of triangle $AFE$ , it is also an altitude of the triangle; combining this with $\angle QIA = x + y$ from the Exterior Angle Theorem gives $\angle FRC = 90 - x - y$ . Also, $\angle VRQ = \angle FRC = 90 - x - y$ because they are vertical angles. This completes part (a).

Solution 2: First we show that the intersection $V'$ of $MP$ with the internal angle bisector of $C$ is the same as the intersection $V''$ of $EF$ with the internal angle bisector of $C.$ Let $D$ denote the intersection of $AB$ with the internal angle bisector of $C,$ and let $a,b,c$ denote the side lengths of $BC, AC, AB.$ By Menelaus on $V'', F, E,$ with respect to $\triangle ADC,$ $\frac{AF}{FD}\cdot \frac{DV''}{V''C}\cdot \frac{CE}{EA}=-1$ $\frac{DV''}{V''C}=-\frac{\frac{bc}{a+b}+\frac{a-b-c}{2}}{\frac{a+b-c}{2}}=\frac{b-a}{b+a}.$ Similarly, $\frac{BP}{PD}\cdot \frac{DV'}{V'C}\cdot \frac{CM}{MB}=-1$ $\frac{DV'}{V'C}=-\frac{\frac{ac}{a+b}-\frac{c}{2}}{\frac{c}{2}}=\frac{b-a}{b+a}.$ Since $V'$ and $V''$ divide $CD$ in the same ratio, they must be the same point. Now, since $\frac{b-a}{b+a}>-1,$ $I$ lies on ray $CV.$ $\blacksquare$

Solution 3: By the Iran Lemma, we know $CI, EF, MP$ concur, so Part $\text{A}$ follows easily.

(b)

Solution 1: Using a similar argument to part (a), point U lies on line $BI$ . Because $\angle MVC = \angle VCA = \angle MCV$ , triangle $VMC$ is isosceles. Similarly, triangle $BMU$ is isosceles, from which we derive that $VM = MC = MB = MU$ . Hence, triangle $VUM$ is isosceles.

Note that $X$ lies on both the circumcircle and the perpendicular bisector of segment $BC$ . Let $D$ be the midpoint of $UV$ ; our goal is to prove that points $X$ , $D$ , and $I$ are collinear, which equates to proving $X$ lies on ray $ID$ .

Because $MD$ is also an altitude of triangle $MVU$ , and $MD$ and $IA$ are both perpendicular to $EF$ , $\overline{MD} \parallel \overline{IA}$ . Furthermore, we have $\angle VMD = \angle UMD = x$ because $APMN$ is a parallelogram. (incomplete)

Solution 2: Let $I_A$ , $I_B$ , and $I_C$ be the excenters of $ABC$ . Note that the circumcircle of $ABC$ is the nine-point circle of $I_AI_BI_C$ . Since $AX$ is the external angle bisector of $\angle BAC$ , $X$ is the midpoint of $I_BI_C$ . Now $UV$ and $I_BI_C$ are parallel since both are perpendicular to the internal angle bisector of $\angle BAC$ . Since $IX$ bisects $I_BI_C$ , it bisects $UV$ as well.

Solution 3: Let $X'$ be the antipode of $X$ with respect to the circumcircle of triangle $ABC$ . Then, by the Incenter-Excenter lemma, $X'$ is the center of a circle containing $B$ , $I$ , and $C$ . Because $XX'$ is a diameter, $XB$ and $XC$ are tangent to the aforementioned circle; thus by a well-known symmedian lemma, $XI$ coincides with the $I$ -symmedian of triangle $IBC$ . From part (a); we know that $BVUC$ is cyclic (we can derive a similar argument for point $U$ ); thus $XI$ coincides with the median of triangle $VIU$ , and we are done.

Solution 4: Let $M_b, M_c$ be the midpoints of arcs $CA, AB$ respectively, and let $T_a$ be the tangency point between the $A$ -mixtilinear incircle of $ABC$ and $\Gamma$ . It's well-known that $T_a \in XI$ , $T_aX$ bisects $M_bM_c$ , $AI \perp EF$ , and $AI \perp M_bM_c$ .

Now, it's easy to see $EF \parallel M_bM_c$ , so $IUV$ and $IM_bM_c$ are homothetic at $I$ . But $T_aX \equiv IX$ bisects $M_bM_c$ , so Part $\text{B}$ follows directly from the homothety. ~ ike.chen

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Difference between revisions of "2014 USAJMO Problems/Problem 6"

Latest revision as of 17:17, 28 August 2021

Problem

Solution

@@ Line 7: / Line 7: @@
 ==Solution==
-'''Extra karma will be awarded to the benefactor who so kindly provides a diagram and additional LATEX for this solution. Before then, please peruse your own diagram.'''
-We will first prove part (a) via contradiction: assume that line <math>IC</math> intersects line <math>MP</math> at Q and line <math>EF</math> and R, with R and Q not equal to V. Let x = <A/2 = <IAE and y = <C/2 = <ICA. We know that <math>MP // AC</math> because <math>MP</math> is a midsegment of triangle <math>ABC</math>; thus, by alternate interior angles (A.I.A) <MVE = <FEA = (180° - 2x) / 2 = 90° - x, because triangle <math>AFE</math> is isosceles. Also by A.I.A, <MQC = <QCA = y. Furthermore, because <math>AI</math> is an angle bisector of triangle <math>AFE</math>, it is also an altitude of the triangle; combining this with <QIA = x + y from the Exterior Angle Theorem gives <FRC = 90 - x - y. <VRQ = <FRC = 90 - x - y because they are vertical angles; however, .... This completes part (a).
+<center>
+<asy>
+unitsize(5cm);
+import olympiad;
+pair A, B, C, I, M, N, P, E, F, U, V, X, R;
+A = dir(190);
+B = dir(120);
+C = dir(350);
+I = incenter(A, B, C);
-Now, we attempt part (b). Using a similar argument to part (a), point U lies on line <math>BI</math>. Because <MVC = <VCA = <MCV, triangle <math>VMC</math> is isosceles. Similarly, triangle <math>BMU</math> is isosceles, from which we derive that VM = MC = MB = MU. Hence, triangle <math>VUM</math> is isosceles.
+label("$A$", A, W);
+label("$B$", B, dir(90));
+label("$C$", C, dir(0));
-Note that X lies on both the circumcircle and the perpendicular bisector of segment <math>BC</math>. Let D be the midpoint of <math>UV</math>; our goal is to prove that points X, D, and I are collinear, which equates to proving X lies on ray <math>ID</math>.
+dot(I); label("$I$", I, SSE);
+draw(A--B--C--cycle);
-Because <math>MD</math> is also an altitude of triangle <math>MVU</math>, and <math>MD</math> and <math>IA</math> are both perpendicular to <math>EF</math>, <math>MD // IA</math>. Furthermore, we have <VMD = <UMD = x because <math>APMN</math> is a parallelogram.
+real r, R;
+r = inradius(A, B, C);
+R = circumradius(A, B, C);
+path G, g;
+G = circumcircle(A, B, C);
+g = incircle(A, B, C);
+draw(G); draw(g);
+label("$\Gamma$", dir(35), dir(35));
+label("$\gamma$", 2/3 * dir(125));
+M = (B+C)/2;
+N = (A+C)/2;
+P = (A+B)/2;
+label("$M$", M, NE);
+label("$N$", N, SE);
+label("$P$", P, W);
+E = tangent(A, I, r, 1);
+F = tangent(A, I, r, 2);
+label("$E$", E, SW);
+label("$F$", F, WNW);
+U = extension(E, F, M, N);
+V = intersectionpoint(P--M, F--E);
+label("$U$", U, S);
+label("$V$", V, NE);
+draw(P--M--U--F);
+X = dir(235);
+label("$X$", X, dir(235));
+draw(X--I, dashed);
+draw(C--V, dashed);
+draw(A--I);
+label("$x^\circ$", A + (0.2,0), dir(90));
+label("$y^\circ$", C + (-0.4,0), dir(90));
+</asy></center>
+(a)
+'''Solution 1:''' We will prove this via contradiction: assume that line <math>IC</math> intersects line <math>MP</math> at <math>Q</math> and line <math>EF</math> and <math>R</math>, with <math>R</math> and <math>Q</math> not equal to <math>V</math>. Let <math>x = \angle A/2 = \angle IAE</math> and <math>y = \angle C/2 = \angle ICA</math>. We know that <math>\overline{MP} \parallel \overline{AC}</math> because <math>MP</math> is a midsegment of triangle <math>ABC</math>; thus, by alternate interior angles (A.I.A) <math>\angle MVE = \angle FEA = (180^\circ - 2x) / 2 = 90^\circ - x</math>, because triangle <math>AFE</math> is isosceles. Also by A.I.A, <math>\angle MQC = \angle QCA = y</math>. Furthermore, because <math>AI</math> is an angle bisector of triangle <math>AFE</math>, it is also an altitude of the triangle; combining this with <math>\angle QIA = x + y</math> from the Exterior Angle Theorem gives <math>\angle FRC = 90 - x - y</math>. Also, <math>\angle VRQ = \angle FRC = 90 - x - y</math> because they are vertical angles. This completes part (a).
+'''Solution 2:''' First we show that the intersection <math>V'</math>of <math>MP</math> with the internal angle bisector of <math>C</math> is the same as the intersection <math>V''</math> of <math>EF</math> with the internal angle bisector of <math>C.</math> Let <math>D</math> denote the intersection of <math>AB</math> with the internal angle bisector of <math>C,</math> and let <math>a,b,c</math> denote the side lengths of <math>BC, AC, AB.</math> By Menelaus on <math>V'', F, E,</math> with respect to <math>\triangle ADC,</math> <cmath>\frac{AF}{FD}\cdot \frac{DV''}{V''C}\cdot \frac{CE}{EA}=-1</cmath>
+<cmath>\frac{DV''}{V''C}=-\frac{\frac{bc}{a+b}+\frac{a-b-c}{2}}{\frac{a+b-c}{2}}=\frac{b-a}{b+a}.</cmath> Similarly, <cmath>\frac{BP}{PD}\cdot \frac{DV'}{V'C}\cdot \frac{CM}{MB}=-1</cmath> <cmath>\frac{DV'}{V'C}=-\frac{\frac{ac}{a+b}-\frac{c}{2}}{\frac{c}{2}}=\frac{b-a}{b+a}.</cmath> Since <math>V'</math> and <math>V''</math> divide <math>CD</math> in the same ratio, they must be the same point. Now, since <math>\frac{b-a}{b+a}>-1,</math> <math>I</math> lies on ray <math>CV.</math> <math>\blacksquare</math>
+'''Solution 3:''' By the Iran Lemma, we know <math>CI, EF, MP</math> concur, so Part <math>\text{A}</math> follows easily.
+(b)
+'''Solution 1:''' Using a similar argument to part (a), point U lies on line <math>BI</math>. Because <math>\angle MVC = \angle VCA = \angle MCV</math>, triangle <math>VMC</math> is isosceles. Similarly, triangle <math>BMU</math> is isosceles, from which we derive that <math>VM = MC = MB = MU</math>. Hence, triangle <math>VUM</math> is isosceles.
+Note that <math>X</math> lies on both the circumcircle and the perpendicular bisector of segment <math>BC</math>. Let <math>D</math> be the midpoint of <math>UV</math>; our goal is to prove that points <math>X</math>, <math>D</math>, and <math>I</math> are collinear, which equates to proving <math>X</math> lies on ray <math>ID</math>.
+Because <math>MD</math> is also an altitude of triangle <math>MVU</math>, and <math>MD</math> and <math>IA</math> are both perpendicular to <math>EF</math>, <math>\overline{MD} \parallel \overline{IA}</math>. Furthermore, we have <math>\angle VMD = \angle UMD = x</math> because <math>APMN</math> is a parallelogram. (incomplete)
+'''Solution 2:''' Let <math>I_A</math>, <math>I_B</math>, and <math>I_C</math> be the excenters of <math>ABC</math>. Note that the circumcircle of <math>ABC</math> is the nine-point circle of <math>I_AI_BI_C</math>. Since <math>AX</math> is the external angle bisector of <math>\angle BAC</math>, <math>X</math> is the midpoint of <math>I_BI_C</math>. Now <math>UV</math> and <math>I_BI_C</math> are parallel since both are perpendicular to the internal angle bisector of <math>\angle BAC</math>. Since <math>IX</math> bisects <math>I_BI_C</math>, it bisects <math>UV</math> as well.
+'''Solution 3:''' Let <math>X'</math> be the antipode of <math>X</math> with respect to the circumcircle of triangle <math>ABC</math>. Then, by the Incenter-Excenter lemma, <math>X'</math> is the center of a circle containing <math>B</math>, <math>I</math>, and <math>C</math>. Because <math>XX'</math> is a diameter, <math>XB</math> and <math>XC</math> are tangent to the aforementioned circle; thus by a well-known symmedian lemma, <math>XI</math> coincides with the <math>I</math>-symmedian of triangle <math>IBC</math>. From part (a); we know that <math>BVUC</math> is cyclic (we can derive a similar argument for point <math>U</math>); thus <math>XI</math> coincides with the median of triangle <math>VIU</math>, and we are done.
+'''Solution 4:''' Let <math>M_b, M_c</math> be the midpoints of arcs <math>CA, AB</math> respectively, and let <math>T_a</math> be the tangency point between the <math>A</math>-mixtilinear incircle of <math>ABC</math> and <math>\Gamma</math>. It's well-known that <math>T_a \in XI</math>, <math>T_aX</math> bisects <math>M_bM_c</math>, <math>AI \perp EF</math>, and <math>AI \perp M_bM_c</math>.
+Now, it's easy to see <math>EF \parallel M_bM_c</math>, so <math>IUV</math> and <math>IM_bM_c</math> are homothetic at <math>I</math>. But <math>T_aX \equiv IX</math> bisects <math>M_bM_c</math>, so Part <math>\text{B}</math> follows directly from the homothety.
+~ ike.chen