Difference between revisions of "2014 USAJMO Problems/Problem 6"
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'''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> | '''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) | ||
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'''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 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 |
Latest revision as of 18:17, 28 August 2021
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