Difference between revisions of "2014 USAJMO Problems/Problem 6"
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− | + | '''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> | <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) | (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>. | 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. | + | 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 | ||
+ | |||
+ | |||
+ | ==Another Solution for Part (a)== | ||
− | + | Let <math>\angle CAI = \angle BAI = x, \angle ACI = \angle BCI = z, \angle ABI = \angle CBI = y</math>. Since <math>AE = AF</math> (they are both tangents), we know that <math>\Delta AEF</math> is isosceles with vertex angle bisected by <math>\overline{AI}</math>. Therefore, <math>\overline{AI} \perp \overline{EF}</math>. This means that <math>\angle AEF = 90 - x \rightarrow \angle CEF = 90 + x</math>. Let <math>V</math> be the intersection of <math>\overleftrightarrow{EF}</math> and <math>\overleftrightarrow{CI}</math>, so <math>\angle EVC = 180 - (90 + x) - z = 90 - x - z = \frac{180-2x-2z}{2}</math> <math>= \frac{2y}{2} = y</math>, so <math>\angle FVI = 180 - y = 180 - \angle FBI</math>, so <math>BFVI</math> is cyclic, meaning <math>\angle BVC = \angle BVI = \angle BFI = 90</math>. Therefore, <math>M</math> is the circumcenter of <math>\Delta BVC</math>, so <math>MV = CM</math>, so <math>\angle CVM = z \rightarrow \angle CMV = 180 - 2z</math>. Then <math>\angle CMV</math> is supplementary to <math>\angle ACB</math>, so <math>\overleftrightarrow{VM} \parallel \overleftrightarrow{AC} \parallel \overleftrightarrow{MP}</math>, so <math>V</math> is on <math>\overleftrightarrow{MP}</math>, so the three lines <math>\overleftrightarrow{EF}, \overleftrightarrow{CI}, \overleftrightarrow{MP}</math> are concurrent at <math>V</math>, meaning that <math>I</math> is on <math>\overrightarrow{CI}</math>, as desired. |
Latest revision as of 19:53, 22 February 2025
Problem
Let be a triangle with incenter
, incircle
and circumcircle
. Let
be the midpoints of sides
,
,
and let
be the tangency points of
with
and
, respectively. Let
be the intersections of line
with line
and line
, respectively, and let
be the midpoint of arc
of
.
(a) Prove that lies on ray
.
(b) Prove that line bisects
.
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]](http://latex.artofproblemsolving.com/3/b/5/3b518f7233f90f68bacb15dc1dcf2b5f1c3e8f21.png)
(a)
Solution 1: We will prove this via contradiction: assume that line intersects line
at
and line
and
, with
and
not equal to
. Let
and
. We know that
because
is a midsegment of triangle
; thus, by alternate interior angles (A.I.A)
, because triangle
is isosceles. Also by A.I.A,
. Furthermore, because
is an angle bisector of triangle
, it is also an altitude of the triangle; combining this with
from the Exterior Angle Theorem gives
. Also,
because they are vertical angles. This completes part (a).
Solution 2: First we show that the intersection of
with the internal angle bisector of
is the same as the intersection
of
with the internal angle bisector of
Let
denote the intersection of
with the internal angle bisector of
and let
denote the side lengths of
By Menelaus on
with respect to
Similarly,
Since
and
divide
in the same ratio, they must be the same point. Now, since
lies on ray
Solution 3: By the Iran Lemma, we know concur, so Part
follows easily.
(b)
Solution 1: Using a similar argument to part (a), point U lies on line . Because
, triangle
is isosceles. Similarly, triangle
is isosceles, from which we derive that
. Hence, triangle
is isosceles.
Note that lies on both the circumcircle and the perpendicular bisector of segment
. Let
be the midpoint of
; our goal is to prove that points
,
, and
are collinear, which equates to proving
lies on ray
.
Because is also an altitude of triangle
, and
and
are both perpendicular to
,
. Furthermore, we have
because
is a parallelogram. (incomplete)
Solution 2: Let ,
, and
be the excenters of
. Note that the circumcircle of
is the nine-point circle of
. Since
is the external angle bisector of
,
is the midpoint of
. Now
and
are parallel since both are perpendicular to the internal angle bisector of
. Since
bisects
, it bisects
as well.
Solution 3: Let be the antipode of
with respect to the circumcircle of triangle
. Then, by the Incenter-Excenter lemma,
is the center of a circle containing
,
, and
. Because
is a diameter,
and
are tangent to the aforementioned circle; thus by a well-known symmedian lemma,
coincides with the
-symmedian of triangle
. From part (a); we know that
is cyclic (we can derive a similar argument for point
); thus
coincides with the median of triangle
, and we are done.
Solution 4: Let be the midpoints of arcs
respectively, and let
be the tangency point between the
-mixtilinear incircle of
and
. It's well-known that
,
bisects
,
, and
.
Now, it's easy to see , so
and
are homothetic at
. But
bisects
, so Part
follows directly from the homothety.
~ ike.chen
Another Solution for Part (a)
Let . Since
(they are both tangents), we know that
is isosceles with vertex angle bisected by
. Therefore,
. This means that
. Let
be the intersection of
and
, so
, so
, so
is cyclic, meaning
. Therefore,
is the circumcenter of
, so
, so
. Then
is supplementary to
, so
, so
is on
, so the three lines
are concurrent at
, meaning that
is on
, as desired.