Difference between revisions of "2016 USAMO Problems/Problem 5"
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Prove that <math>\overline{OI}</math> is parallel to <math>\ell,</math> where <math>O</math> is the circumcenter of triangle <math>ABC,</math> and <math>I</math> is the incenter of triangle <math>ABC.</math> | Prove that <math>\overline{OI}</math> is parallel to <math>\ell,</math> where <math>O</math> is the circumcenter of triangle <math>ABC,</math> and <math>I</math> is the incenter of triangle <math>ABC.</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | {{ | + | |
+ | Let <math>D</math> be the intersection of line <math>AI</math> and the circumcircle of <math>\Delta ABC</math> (other than <math>A</math>), then <math>OD\perp BC</math>. Let <math>R</math> be the point such that <math>NPQR</math> is a rhombus. It follows that <math>OD\perp QR</math>. | ||
+ | |||
+ | Since <math>AM=AQ</math>, <math>AI\perp MQ</math>, or <math>DI\perp MQ</math>. It follows that <math>\angle ODI=\angle RQM</math>. | ||
+ | |||
+ | Since <math>BO=OD</math>, <math>MA=AQ</math>, <math>\angle BOD=\angle MAQ</math>, it follows that <math>\Delta BOD\sim\Delta MAQ</math>, so <math>AQ/MQ=OD/BD</math>. | ||
+ | |||
+ | It is given that <math>AQ=NP=RQ</math>, and by basic properties of the incenter, <math>ID=BD</math>. Therefore, <math>RQ/MQ=OD/ID</math>, so <math>\Delta RQM\sim\Delta ODI</math>. | ||
+ | Since the rotation between the two triangles in 90 degrees, <math>OI\perp MR</math>. However, <math>l</math> is parallel to the bisector of <math>MNR</math>, which is perpendicular to <math>MR</math>, so we are done. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Write <math>\angle{JKL} = K</math> for all <math>J,K,L</math> chosen as distinct vertices of triangle <math>ABC</math>. Define <math>a, b, c</math> as sides opposite to angles <math>A, B</math>, and <math>C</math>, respectively. Place the triangle in the Euclidean plane with <math>A</math> at the origin and <math>C</math> on the positive x-axis. Assume without loss of generality that C is acute. | ||
+ | |||
+ | Consider the sides of the pentagon as vectors and note that <cmath>\overrightarrow{AM} + \overrightarrow{MN} + \overrightarrow{NP} = \overrightarrow{AQ} + \overrightarrow{QP} \qquad (1)</cmath> | ||
+ | |||
+ | Define <math>\delta</math> and <math>\gamma</math> as the angles made between the positive x-axis and <math>\overrightarrow{MN}</math> and <math>\overrightarrow{QP}</math>, respectively. Considering the x and y coordinates of the vectors in <math>(1)</math>, it follows that | ||
+ | <cmath>\cos \delta - \cos \gamma = 1 - \cos A - \cos C \qquad (2)</cmath> | ||
+ | <cmath>\sin \delta - \sin \gamma = \sin C - \sin A \qquad (3)</cmath> | ||
+ | |||
+ | Suppose <math>\sin C - \sin A = 0</math>. Then <math>A = C</math>, and the triangle is isosceles. In this case, it is clear by symmetry that <math>\overline{OI}</math> is vertical. Further, since point <math>S</math> exists, <math>\delta \neq \gamma</math>, so <math>\delta + \gamma = 180</math> and <math>\overrightarrow{MN} + \overrightarrow{QP}</math> must be vertical as well. | ||
+ | |||
+ | For the remainder of the proof, assume <math>\sin C \neq \sin A</math>. Note that <cmath>\frac{\cos y – \cos x}{\sin x - \sin y} = \frac{ (\cos y - \cos x)(\sin x + \sin y)}{ (\sin x - \sin y)(\sin x + \sin y)} = \frac {\sin x + \sin y}{\cos x + \cos y} </cmath> whenever <math>x, y \in \mathbb{R}</math> and <math>\sin x \neq \sin y</math>. Note further that the slope of the line defined by the vector formed by summing vectors <math>(\cos x, \sin x)</math> and <math>(\cos y, \sin y)</math> is this expression. Since <math>\ell</math> is parallel to <math>\overrightarrow{MN} + \overrightarrow{QP}</math>, the slope of <math>\ell</math> can be formed by dividing expressions in <math>(2)</math> and <math>(3)</math> and inverting the sign: <cmath>\frac{\cos A + \cos C – 1}{\sin C - \sin A} \qquad (4)</cmath> | ||
+ | |||
+ | Determine the coordinates of <math>I</math> by drawing perpendiculars from <math>I</math> to the sides and vertices of the triangle. By exploiting congruence between pairs of right triangles that share a vertex, one can partition <math>a,b,c</math> into <math>p+q, p+r, q+r</math> where <math>p,q,r</math> are bases of these triangles that lie on the sides of triangle <math>ABC</math>. From here it is clear that <math>I = \left(\frac{b + c – a}{2} , \frac{b + c – a}{2}\tan(A / 2)\right)</math>. | ||
+ | |||
+ | To find the coordinates of <math>O</math>, note that <math>\angle{OJK} + \angle{OJL} = \angle{J}</math> and that <math>\angle {OJK} = \angle{OKJ}</math> in any acute triangle <math>JKL</math>. It easily follows that <math>\angle{OJK} = 90 – L</math>. Note also that the perpendicular from <math>O</math> to <math>\overline{JK}</math> bisects <math>\overline{JK}</math>. Hence, <cmath>O = \left(\frac{b}{2},\frac{b}{2} \cot B\right) \qquad (5)</cmath> if triangle <math>ABC</math> is acute. | ||
+ | |||
+ | If triangle <math>JKL</math> is obtuse at <math>\angle J</math>, then it can be similarly shown that <math>\angle{OKL} = \angle{OLK} = J – 90</math> but that the remaining angles of this form are still <math>90-L</math> and <math>90-K</math>. It easily follows that <math>(5)</math> holds if <math>\angle A</math> is obtuse. If <math>\angle B</math> is obtuse then <math>\angle OAC = B – 90</math> and the <math>y</math> coordinate of <math>O</math> is <math>-\frac{b}{2} \tan{B-90}</math>. From this, <math>(5)</math> follows in this case as well. | ||
+ | |||
+ | We can conclude the slope of <math>\overline {OI}</math> is <cmath>\left(\frac{b \cot {B} – (b + c – a) \tan(A/2)}{a – c}\right) = \left( \frac {\cos B – \tan{(A/2)}\sin B}{\sin A - \sin C}\right) + \tan(A/2) \quad (6)</cmath> by the Law of Sines and rearrangement. | ||
+ | |||
+ | Setting <math>(6) = (4)</math> is equivalent to <cmath>1 - \cos A - \cos C = \cos B + \tan(A/2)(\sin A - \sin B - \sin C)</cmath> | ||
+ | |||
+ | Since <math>\tan(A/2) = \frac{\sin A}{1 + \cos A}</math>, this equation is equivalent to <cmath>(1 + \cos A)(\cos A + \cos B + \cos C – 1) = (\sin A)(\sin B + \sin C - \sin A)</cmath> | ||
+ | |||
+ | This equation is equivalent to <cmath>\cos(A + B) + \cos(A + C) + \cos B + \cos C = 0</cmath> which is evident. | ||
+ | |||
+ | ==Solution 3== | ||
+ | [[File:2016 USAMO 5a.png|500px|right]] | ||
+ | Let <math>A', B',</math> and <math>C'</math> be the arc midpoints of <math>BC, CA, AB,</math> respectively. Let <math>E</math> be crosspoint of <math>AI</math> and <math>B'C'.</math> | ||
+ | |||
+ | Therefore <math>O</math> is the circumcenter of triangle <math>A'B'C'.</math> | ||
+ | |||
+ | Points <math>A, I, E,</math> and <math>A'</math> are collinear. | ||
+ | <math>\angle A'EB' = \frac {\overset{\Large\frown} {AC'}+\overset{\Large\frown} {B'C} +\overset{\Large\frown} {CA'}}{2} = 90^\circ \implies AA' \perp B'C'</math> | ||
+ | <math>\implies I</math> is orthocenter of <math>\triangle A'B'C' \implies</math> | ||
+ | |||
+ | <math>IO</math> is the Euler line of <math>\triangle A'B'C'.</math> | ||
+ | |||
+ | Let <math>G</math> be the centroid of <math>\triangle A'B'C' \implies G </math> lies on line <math>IO</math> | ||
+ | <math>\implies \overline {OG} = \frac {\overline{OA'} + \overline{OB'} + \overline{OC'}}{3}</math> is paraller to <math>\overline{OI}.</math> | ||
+ | <math>\overline{OB'} \perp \overline{AC} \implies \overline{OB'} \perp \overline{QA}.</math> | ||
+ | [[File:2016 USAMO 5.png|250px|right]] | ||
+ | Similarly <math>\overline{OC'} \perp \overline{AM}, \overline{OA'} \perp \overline{NP},</math> rotation from <math>\overline{OA'}</math> to <math>\overline{NP}</math>, from <math>\overline{OB'}</math> to <math>\overline{QA},</math> and from <math>\overline{OC'}</math> to <math>\overline{AM}</math> is in clockwise direction, <math>|\overline{QA}|=| \overline{AM}| =|\overline{NP}|, |\overline{OA'}| = |\overline{OB'}| = |\overline{OC'}| \implies</math> | ||
+ | |||
+ | <math>\overline{QA} + \overline{AM} + \overline{NP} = \overline{QM} + \overline{NP}</math> is perpendicular to <math>|\overline{OI}.</math> | ||
+ | |||
+ | <math>|\overline{MN}| = |\overline{QP}|</math> therefore in accordance with <i><b>Claim</b></i> <math>\overline{MN} + \overline{QP}</math> is parallel to <math>\overline{OI}.</math> | ||
+ | |||
+ | This sum is parallel to <math>\ell,</math>, so we are done. | ||
+ | |||
+ | <i><b>Claim</b></i> | ||
+ | |||
+ | Let <math>|\overline{BA}|= |\overline{CD}|, \overline{CB} + |\overline{BA} + |\overline{AD} = |\overline{CD}.</math> | ||
+ | Then <math>(\overline{BA} + \overline{CD}) \perp (\overline{CB} + \overline{AD}).</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>(\overline{BA} + \overline{CD}) \cdot (\overline{CB} + \overline{AD}) = (\overline{BA} + \overline{CD}) \cdot (\overline{CB} + \overline{CD} – \overline{CB} – \overline{BA}) = |\overline{CD}|^2 – |\overline{BA}|^2 = 0.</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | |||
+ | ==Video Solution by MOP 2024== | ||
+ | https://youtu.be/OnGdqIvOBd4 | ||
+ | |||
+ | ~r00tsOfUnity | ||
− | |||
==See also== | ==See also== | ||
{{USAMO newbox|year=2016|num-b=4|num-a=6}} | {{USAMO newbox|year=2016|num-b=4|num-a=6}} | ||
+ | {{MAA Notice}} |
Latest revision as of 00:09, 12 October 2023
Problem
An equilateral pentagon is inscribed in triangle
such that
and
Let
be the intersection of lines
and
Denote by
the angle bisector of
Prove that is parallel to
where
is the circumcenter of triangle
and
is the incenter of triangle
Solution 1
Let be the intersection of line
and the circumcircle of
(other than
), then
. Let
be the point such that
is a rhombus. It follows that
.
Since ,
, or
. It follows that
.
Since ,
,
, it follows that
, so
.
It is given that , and by basic properties of the incenter,
. Therefore,
, so
.
Since the rotation between the two triangles in 90 degrees,
. However,
is parallel to the bisector of
, which is perpendicular to
, so we are done.
Solution 2
Write for all
chosen as distinct vertices of triangle
. Define
as sides opposite to angles
, and
, respectively. Place the triangle in the Euclidean plane with
at the origin and
on the positive x-axis. Assume without loss of generality that C is acute.
Consider the sides of the pentagon as vectors and note that
Define and
as the angles made between the positive x-axis and
and
, respectively. Considering the x and y coordinates of the vectors in
, it follows that
Suppose . Then
, and the triangle is isosceles. In this case, it is clear by symmetry that
is vertical. Further, since point
exists,
, so
and
must be vertical as well.
For the remainder of the proof, assume . Note that
whenever
and
. Note further that the slope of the line defined by the vector formed by summing vectors
and
is this expression. Since
is parallel to
, the slope of
can be formed by dividing expressions in
and
and inverting the sign:
Determine the coordinates of by drawing perpendiculars from
to the sides and vertices of the triangle. By exploiting congruence between pairs of right triangles that share a vertex, one can partition
into
where
are bases of these triangles that lie on the sides of triangle
. From here it is clear that
.
To find the coordinates of , note that
and that
in any acute triangle
. It easily follows that
. Note also that the perpendicular from
to
bisects
. Hence,
if triangle
is acute.
If triangle is obtuse at
, then it can be similarly shown that
but that the remaining angles of this form are still
and
. It easily follows that
holds if
is obtuse. If
is obtuse then
and the
coordinate of
is
. From this,
follows in this case as well.
We can conclude the slope of is
by the Law of Sines and rearrangement.
Setting is equivalent to
Since , this equation is equivalent to
This equation is equivalent to which is evident.
Solution 3
Let and
be the arc midpoints of
respectively. Let
be crosspoint of
and
Therefore is the circumcenter of triangle
Points and
are collinear.
is orthocenter of
is the Euler line of
Let be the centroid of
lies on line
is paraller to
Similarly rotation from
to
, from
to
and from
to
is in clockwise direction,
is perpendicular to
therefore in accordance with Claim
is parallel to
This sum is parallel to , so we are done.
Claim
Let
Then
Proof
vladimir.shelomovskii@gmail.com, vvsss
Video Solution by MOP 2024
~r00tsOfUnity
See also
2016 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.