Difference between revisions of "2013 USAMO Problems/Problem 1"
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==Problem== | ==Problem== | ||
− | In triangle <math>ABC</math>, points <math>P,Q,R</math> lie on sides <math>BC,CA,AB</math> respectively. Let <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> denote the circumcircles of triangles <math>AQR</math>, <math>BRP</math>, <math>CPQ</math>, respectively. Given the fact that segment <math>AP</math> intersects <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> again at <math>X,Y,Z</math> respectively, prove that <math>YX/XZ=BP/PC</math> | + | In triangle <math>ABC</math>, points <math>P,Q,R</math> lie on sides <math>BC,CA,AB</math> respectively. Let <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> denote the circumcircles of triangles <math>AQR</math>, <math>BRP</math>, <math>CPQ</math>, respectively. Given the fact that segment <math>AP</math> intersects <math>\omega_A</math>, <math>\omega_B</math>, <math>\omega_C</math> again at <math>X,Y,Z</math> respectively, prove that <math>YX/XZ=BP/PC</math>. |
==Solution 1== | ==Solution 1== | ||
Line 81: | Line 81: | ||
In this solution, all lengths and angles are directed. | In this solution, all lengths and angles are directed. | ||
− | Firstly, it is easy to see by that <math>\omega_A, \omega_B, \omega_C</math> concur at a point <math>M</math>. Let <math>XM</math> meet <math>\omega_B, \omega_C</math> again at <math>D</math> and <math>E</math>, respectively. Then by Power of a Point, we have <cmath>XM \cdot XE = XZ \cdot XP \quad\text{and}\quad XM \cdot XD = XY \cdot XP</cmath> Thusly <cmath>\frac{XY}{XZ} = \frac{XD}{XE}</cmath> But we claim that <math>\triangle XDP \sim \triangle PBM</math>. Indeed, <cmath>\measuredangle XDP = \measuredangle MDP = \measuredangle MBP = - \measuredangle PBM</cmath> and <cmath>\measuredangle DXP = \measuredangle MXY = \measuredangle MXA = \measuredangle MRA = 180^\circ - \measuredangle MRB = \measuredangle MPB = -\measuredangle BPM</cmath> | + | Firstly, it is easy to see by that <math>\omega_A, \omega_B, \omega_C</math> concur at a point <math>M</math> (the Miquel point). Let <math>XM</math> meet <math>\omega_B, \omega_C</math> again at <math>D</math> and <math>E</math>, respectively. Then by Power of a Point, we have <cmath>XM \cdot XE = XZ \cdot XP \quad\text{and}\quad XM \cdot XD = XY \cdot XP</cmath> Thusly <cmath>\frac{XY}{XZ} = \frac{XD}{XE}</cmath> But we claim that <math>\triangle XDP \sim \triangle PBM</math>. Indeed, <cmath>\measuredangle XDP = \measuredangle MDP = \measuredangle MBP = - \measuredangle PBM</cmath> and <cmath>\measuredangle DXP = \measuredangle MXY = \measuredangle MXA = \measuredangle MRA = 180^\circ - \measuredangle MRB = \measuredangle MPB = -\measuredangle BPM</cmath> |
Therefore, <math>\frac{XD}{XP} = \frac{PB}{PM}</math>. Analogously we find that <math>\frac{XE}{XP} = \frac{PC}{PM}</math> and we are done. | Therefore, <math>\frac{XD}{XP} = \frac{PB}{PM}</math>. Analogously we find that <math>\frac{XE}{XP} = \frac{PC}{PM}</math> and we are done. | ||
− | courtesy v_enhance | + | courtesy v_enhance, minor clarification by integralarefun |
---- | ---- | ||
Line 113: | Line 113: | ||
==Solution 4== | ==Solution 4== | ||
− | We | + | We will use some construction arguments to solve the problem. Let <math>\angle BAC=\alpha,</math> <math>\angle ABC=\beta,</math> <math>\angle ACB=\gamma,</math> and let <math>\angle APB=\theta.</math> We construct lines through the points <math>Q,</math> and <math>R</math> that intersect with <math>\triangle ABC</math> at the points <math>Q</math> and <math>R,</math> respectively, and that intersect each other at <math>T.</math> We will construct these lines such that <math>\angle CQV=\angle ARV=\theta.</math> |
− | |||
+ | Now we let the intersections of <math>AP</math> with <math>RV</math> and <math>QU</math> be <math>Y'</math> and <math>Z',</math> respectively. This construction is as follows. | ||
+ | <asy> | ||
+ | import graph; size(12cm); | ||
+ | real labelscalefactor = 1.9; | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); | ||
+ | pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); | ||
+ | |||
+ | draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle); | ||
+ | draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); | ||
+ | draw((-7.61,-5)--(7.09,-5)); | ||
+ | draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); | ||
+ | draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); | ||
+ | draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); | ||
+ | draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5)); | ||
+ | dot((-3.6988888888888977,6.426666666666669)); | ||
+ | label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); | ||
+ | dot((-7.61,-5)); | ||
+ | label("$B$", (-7.61,-5), SW * labelscalefactor); | ||
+ | dot((7.09,-5)); | ||
+ | label("$C$", (7.09,-5), SE * labelscalefactor); | ||
+ | dot((-2.958888888888898,-5)); | ||
+ | label("$P$", (-2.958888888888898,-5), S * labelscalefactor); | ||
+ | dot((0.5968131669050584,1.8770271258031248)); | ||
+ | label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); | ||
+ | dot((-5.053354907372894,2.4694710603912564)); | ||
+ | label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); | ||
+ | dot((-3.143912404905382,-2.142970212141873)); | ||
+ | label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); | ||
+ | dot((-3.413789986031826,2.0243286531799747)); | ||
+ | label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); | ||
+ | dot((-3.3284001481939356,0.7057864725120093)); | ||
+ | label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); | ||
+ | dot((1.7922953932137468,0.6108747864253139)); | ||
+ | label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); | ||
+ | dot((-5.8024625203461,-5)); | ||
+ | label("$U$", (-5.8024625203461,-5), S * labelscalefactor); | ||
+ | dot((-0.10264330299819162,1.125351256231488)); | ||
+ | label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); | ||
+ | </asy> | ||
+ | |||
+ | We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have, | ||
+ | <cmath>\begin{align*} | ||
+ | \angle BRY'+\angle BPY' | ||
+ | &=180^\circ-\theta+\theta\\ | ||
+ | &=180^\circ. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Since the opposite angles of quadrilateral <math>RY'PB</math> add up to <math>180^\circ,</math> it must be cyclic. Similarly, we can also show that quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic. | ||
+ | |||
+ | Since points <math>Y'</math> and <math>Z'</math> lie on <math>AP,</math> we know that, | ||
+ | <cmath>Y'=\omega_B\cap AP</cmath> | ||
+ | and that | ||
+ | <cmath>Z'=\omega_C\cap AP.</cmath> | ||
+ | |||
+ | Hence, the points <math>Y'</math> and <math>Z'</math> coincide with the given points <math>Y</math> and <math>Z,</math> respectively. | ||
+ | |||
+ | Since quadrilateral <math>AQTR</math> is also cyclic, we have, | ||
+ | <cmath>\begin{align*} | ||
+ | \angle Y'TZ' | ||
+ | &=180^\circ-\angle RTQ\\ | ||
+ | &=180^\circ-(180^\circ-\angle RAQ)\\ | ||
+ | &=\angle RAQ\\ | ||
+ | &=\alpha. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Similarly, since quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic, we have, | ||
+ | <cmath>\begin{align*} | ||
+ | \angle TY'Z' | ||
+ | &=180^\circ-\angle RY'P\\ | ||
+ | &=180^\circ-(180^\circ-\angle RBP)\\ | ||
+ | &=\angle RBP\\ | ||
+ | &=\beta, | ||
+ | \end{align*}</cmath> | ||
+ | and, | ||
+ | <cmath>\begin{align*} | ||
+ | \angle Y'Z'T | ||
+ | &=180^\circ-\angle PZ'Q\\ | ||
+ | &=180^\circ-(180^\circ-\angle PCQ)\\ | ||
+ | &=\angle PCQ\\ | ||
+ | &=\gamma. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Since these three angles are of <math>\triangle TY'Z',</math> and they are equal to corresponding angles of <math>\triangle ABC,</math> by AA similarity, we know that <math>\triangle TY'Z'\sim \triangle ABC.</math> | ||
+ | |||
+ | We now consider the point <math>X=\omega_c\cap AC.</math> We know that the points <math>A,</math> <math>Q,</math> <math>T,</math> and <math>R</math> are concyclic. Hence, the points <math>A,</math> <math>T,</math> <math>X,</math> and <math>R</math> must also be concyclic. | ||
+ | |||
+ | Hence, quadrilateral <math>AQTX</math> is cyclic. | ||
− | |||
<asy> | <asy> | ||
import graph; size(12cm); | import graph; size(12cm); | ||
− | real labelscalefactor = | + | real labelscalefactor = 1.9; |
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); | ||
− | |||
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); | pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); | ||
Line 133: | Line 217: | ||
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); | draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); | ||
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5)); | draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5)); | ||
+ | draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488)); | ||
+ | draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564)); | ||
+ | draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488)); | ||
dot((-3.6988888888888977,6.426666666666669)); | dot((-3.6988888888888977,6.426666666666669)); | ||
− | label("$A$", (-3. | + | label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); |
dot((-7.61,-5)); | dot((-7.61,-5)); | ||
− | label("$B$", (-7. | + | label("$B$", (-7.61,-5), SW * labelscalefactor); |
dot((7.09,-5)); | dot((7.09,-5)); | ||
− | label("$C$", (7. | + | label("$C$", (7.09,-5), SE * labelscalefactor); |
dot((-2.958888888888898,-5)); | dot((-2.958888888888898,-5)); | ||
− | label("$P$", (-2. | + | label("$P$", (-2.958888888888898,-5), S * labelscalefactor); |
dot((0.5968131669050584,1.8770271258031248)); | dot((0.5968131669050584,1.8770271258031248)); | ||
− | label("$Q$", (0. | + | label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); |
dot((-5.053354907372894,2.4694710603912564)); | dot((-5.053354907372894,2.4694710603912564)); | ||
− | label("$R$", (- | + | label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); |
dot((-3.143912404905382,-2.142970212141873)); | dot((-3.143912404905382,-2.142970212141873)); | ||
− | label("$Z'$", (-3. | + | label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); |
dot((-3.413789986031826,2.0243286531799747)); | dot((-3.413789986031826,2.0243286531799747)); | ||
− | label("$Y'$", (-3. | + | label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); |
dot((-3.3284001481939356,0.7057864725120093)); | dot((-3.3284001481939356,0.7057864725120093)); | ||
− | label("$X | + | label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); |
dot((1.7922953932137468,0.6108747864253139)); | dot((1.7922953932137468,0.6108747864253139)); | ||
− | label("$V$", (1. | + | label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); |
dot((-5.8024625203461,-5)); | dot((-5.8024625203461,-5)); | ||
− | label("$U$", (-5. | + | label("$U$", (-5.8024625203461,-5), S * labelscalefactor); |
dot((-0.10264330299819162,1.125351256231488)); | dot((-0.10264330299819162,1.125351256231488)); | ||
− | label("$T$", (-0. | + | label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); |
</asy> | </asy> | ||
+ | |||
+ | Since the angles <math>\angle ART</math> and <math>\angle AXT</math> are inscribed in the same arc <math>\overarc{AT},</math> we have, | ||
+ | <cmath>\begin{align*} | ||
+ | \angle AXT | ||
+ | &=\angle ART\\ | ||
+ | &=\theta. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | Consider by this result, we can deduce that the homothety that maps <math>ABC</math> to <math>TY'Z'</math> will map <math>P</math> to <math>X.</math> Hence, we have that, | ||
+ | <cmath>Y'X/XZ'=BP/PC.</cmath> | ||
+ | |||
+ | Since <math>Y'=Y</math> and <math>Z'=Z</math> hence, | ||
+ | <cmath>YX/XZ=BP/PC,</cmath> | ||
+ | |||
+ | as required. | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 18:22, 10 May 2023
Problem
In triangle , points
lie on sides
respectively. Let
,
,
denote the circumcircles of triangles
,
,
, respectively. Given the fact that segment
intersects
,
,
again at
respectively, prove that
.
Solution 1
In this solution, all lengths and angles are directed.
Firstly, it is easy to see by that concur at a point
(the Miquel point). Let
meet
again at
and
, respectively. Then by Power of a Point, we have
Thusly
But we claim that
. Indeed,
and
Therefore,
. Analogously we find that
and we are done.
courtesy v_enhance, minor clarification by integralarefun
Solution 2
Diagram Refer to the Diagram link.
By Miquel's Theorem, there exists a point at which intersect. We denote this point by
Now, we angle chase:
In addition, we have
Now, by the Ratio Lemma, we have
(by the Law of Sines in
)
(by the Law of Sines in
)
by the Ratio Lemma.
The proof is complete.
Solution 3
Use directed angles modulo .
Lemma.
Proof.
Now, it follows that (now not using directed angles)
using the facts that
and
,
and
are similar triangles, and that
equals twice the circumradius of the circumcircle of
.
Solution 4
We will use some construction arguments to solve the problem. Let
and let
We construct lines through the points
and
that intersect with
at the points
and
respectively, and that intersect each other at
We will construct these lines such that
Now we let the intersections of with
and
be
and
respectively. This construction is as follows.
We know that Hence, we have,
Since the opposite angles of quadrilateral add up to
it must be cyclic. Similarly, we can also show that quadrilaterals
and
are also cyclic.
Since points and
lie on
we know that,
and that
Hence, the points and
coincide with the given points
and
respectively.
Since quadrilateral is also cyclic, we have,
Similarly, since quadrilaterals and
are also cyclic, we have,
and,
Since these three angles are of and they are equal to corresponding angles of
by AA similarity, we know that
We now consider the point We know that the points
and
are concyclic. Hence, the points
and
must also be concyclic.
Hence, quadrilateral is cyclic.
Since the angles and
are inscribed in the same arc
we have,
Consider by this result, we can deduce that the homothety that maps to
will map
to
Hence, we have that,
Since and
hence,
as required.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.