https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Negia&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T04:43:22ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1639282013 USAMO Problems/Problem 12021-10-23T02:03:51Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
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><br />
<br />
<br />
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.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,<br />
<cmath>\begin{align*}<br />
\angle BRY'+\angle BPY'<br />
&=180^\circ-\theta+\theta\\<br />
&=180^\circ.<br />
\end{align*}</cmath><br />
<br />
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.<br />
<br />
Since points <math>Y'</math> and <math>Z'</math> lie on <math>AP,</math> we know that,<br />
<cmath>Y'=\omega_B\cap AP</cmath><br />
and that<br />
<cmath>Z'=\omega_C\cap AP.</cmath><br />
<br />
Hence, the points <math>Y'</math> and <math>Z'</math> coincide with the given points <math>Y</math> and <math>Z,</math> respectively.<br />
<br />
Since quadrilateral <math>AQTR</math> is also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle Y'TZ'<br />
&=180^\circ-\angle RTQ\\<br />
&=180^\circ-(180^\circ-\angle RAQ)\\<br />
&=\angle RAQ\\<br />
&=\alpha.<br />
\end{align*}</cmath><br />
<br />
Similarly, since quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle TY'Z'<br />
&=180^\circ-\angle RY'P\\<br />
&=180^\circ-(180^\circ-\angle RBP)\\<br />
&=\angle RBP\\<br />
&=\beta,<br />
\end{align*}</cmath><br />
and,<br />
<cmath>\begin{align*}<br />
\angle Y'Z'T<br />
&=180^\circ-\angle PZ'Q\\<br />
&=180^\circ-(180^\circ-\angle PCQ)\\<br />
&=\angle PCQ\\<br />
&=\gamma.<br />
\end{align*}</cmath><br />
<br />
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><br />
<br />
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.<br />
<br />
Hence, quadrilateral <math>AQTX</math> is cyclic.<br />
<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564));<br />
draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
Since the angles <math>\angle ART</math> and <math>\angle AXT</math> are inscribed in the same arc <math>\overarc{AT},</math> we have,<br />
<cmath>\begin{align*}<br />
\angle AXT<br />
&=\angle ART\\<br />
&=\theta.<br />
\end{align*}</cmath><br />
<br />
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,<br />
<cmath>Y'X/XZ'=BP/PC.</cmath><br />
<br />
Since <math>Y'=Y</math> and <math>Z'=Z</math> hence,<br />
<cmath>YX/XZ=BP/PC,</cmath><br />
<br />
as required.<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1639272013 USAMO Problems/Problem 12021-10-23T02:02:47Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We will use some construction arguments to solve the problem. (The diagrams are variations of Evan Chen's diagrams)<br />
<br />
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><br />
<br />
<br />
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.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,<br />
<cmath>\begin{align*}<br />
\angle BRY'+\angle BPY'<br />
&=180^\circ-\theta+\theta\\<br />
&=180^\circ.<br />
\end{align*}</cmath><br />
<br />
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.<br />
<br />
Since points <math>Y'</math> and <math>Z'</math> lie on <math>AP,</math> we know that,<br />
<cmath>Y'=\omega_B\cap AP</cmath><br />
and that<br />
<cmath>Z'=\omega_C\cap AP.</cmath><br />
<br />
Hence, the points <math>Y'</math> and <math>Z'</math> coincide with the given points <math>Y</math> and <math>Z,</math> respectively.<br />
<br />
Since quadrilateral <math>AQTR</math> is also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle Y'TZ'<br />
&=180^\circ-\angle RTQ\\<br />
&=180^\circ-(180^\circ-\angle RAQ)\\<br />
&=\angle RAQ\\<br />
&=\alpha.<br />
\end{align*}</cmath><br />
<br />
Similarly, since quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle TY'Z'<br />
&=180^\circ-\angle RY'P\\<br />
&=180^\circ-(180^\circ-\angle RBP)\\<br />
&=\angle RBP\\<br />
&=\beta,<br />
\end{align*}</cmath><br />
and,<br />
<cmath>\begin{align*}<br />
\angle Y'Z'T<br />
&=180^\circ-\angle PZ'Q\\<br />
&=180^\circ-(180^\circ-\angle PCQ)\\<br />
&=\angle PCQ\\<br />
&=\gamma.<br />
\end{align*}</cmath><br />
<br />
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><br />
<br />
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.<br />
<br />
Hence, quadrilateral <math>AQTX</math> is cyclic.<br />
<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564));<br />
draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
Since the angles <math>\angle ART</math> and <math>\angle AXT</math> are inscribed in the same arc <math>\overarc{AT},</math> we have,<br />
<cmath>\begin{align*}<br />
\angle AXT<br />
&=\angle ART\\<br />
&=\theta.<br />
\end{align*}</cmath><br />
<br />
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,<br />
<cmath>Y'X/XZ'=BP/PC.</cmath><br />
<br />
Since <math>Y'=Y</math> and <math>Z'=Z</math> hence,<br />
<cmath>YX/XZ=BP/PC,</cmath><br />
<br />
as required.<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Problems&diff=1516252001 IMO Problems2021-04-16T01:52:17Z<p>Negia: /* Problem 6 */</p>
<hr />
<div>==Problem 1==<br />
<br />
Consider an acute triangle <math>\triangle ABC</math>. Let <math>P</math> be the foot of the altitude of triangle <math>\triangle ABC</math> issuing from the vertex <math>A</math>, and let <math>O</math> be the [[circumcenter]] of triangle <math>\triangle ABC</math>. Assume that <math>\angle C \geq \angle B+30^{\circ}</math>. Prove that <math>\angle A+\angle COP < 90^{\circ}</math>.<br />
<br />
==Problem 2==<br />
<br />
Let <math>a,b,c</math> be positive real numbers. Prove that <math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1</math>.<br />
<br />
==Problem 3==<br />
<br />
Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.<br />
<br />
==Problem 4==<br />
<br />
Let <math>n_1, n_2, \dots , n_m</math> be integers where <math>m>1</math> is odd. Let <math>x = (x_1, \dots , x_m)</math> denote a permutation of the integers <math>1, 2, \cdots , m</math>. Let <math>f(x) = x_1n_1 + x_2n_2 + ... + x_mn_m</math>. Show that for some distinct permutations <math>a</math>, <math>b</math> the difference <math>f(a) - f(b)</math> is a multiple of <math>m!</math>.<br />
<br />
==Problem 5==<br />
<br />
<math>ABC</math> is a [[triangle]]. <math>X</math> lies on <math>BC</math> and <math>AX</math> bisects [[angle]] <math>A</math>. <math>Y</math> lies on <math>CA</math> and <math>BY</math> bisects angle <math>B</math>. Angle <math>A</math> is <math>60^{\circ}</math>. <math>AB + BX = AY + YB</math>. Find all possible values for angle <math>B</math>.<br />
<br />
==Problem 6==<br />
<br />
<math>k > l > m > n</math> are positive integers such that <math>km + ln = (k+l-m+n)(-k+l+m+n)</math>. Prove that <math>kl+mn</math> is not prime.<br />
<br />
==See Also==<br />
<br />
* [[2001 IMO]]<br />
* [[IMO Problems and Solutions]]<br />
* [[IMO]]</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2006_IMO_Problems&diff=1514182006 IMO Problems2021-04-13T03:18:08Z<p>Negia: /* Problem 6 */</p>
<hr />
<div>==Problem 1==<br />
Let <math>ABC</math> be a triangle with incentre <math>I.</math> A point <math>P</math> in the interior of the triangle satisfies <math>\angle PBA + \angle PCA = \angle PBC + \angle PCB</math>.<br />
Show that <math>AP \ge AI,</math> and that equality holds if and only if <math>P = I.</math><br />
<br />
==Problem 2==<br />
Let <math>P</math> be a regular 2006 sided polygon. A diagonal of <math>P</math> is called good if its endpoints divide the boundary of <math>P</math> into two parts, each composed of an odd number of sides of <math>P</math>. The sides of <math>P</math> are also called good. Suppose <math>P</math> has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of <math>P</math>. Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.<br />
<br />
==Problem 3==<br />
Determine the least real number <math>M</math> such that the inequality <cmath> \left| ab\left(a^{2}-b^{2}\right)+bc\left(b^{2}-c^{2}\right)+ca\left(c^{2}-a^{2}\right)\right|\leq M\left(a^{2}+b^{2}+c^{2}\right)^{2} </cmath> holds for all real numbers <math>a,b</math> and <math>c</math><br />
<br />
==Problem 4==<br />
Determine all pairs <math>(x, y)</math> of integers such that <cmath>1+2^{x}+2^{2x+1}= y^{2}.</cmath><br />
<br />
==Problem 5==<br />
Let <math>P(x)</math> be a polynomial of degree <math>n>1</math> with integer coefficients, and let <math>k</math> be a positive integer. Consider the polynomial <math>Q(x) = P( P ( \ldots P(P(x)) \ldots ))</math>, where <math>P</math> occurs <math>k</math> times. Prove that there are at most <math>n</math> integers <math>t</math> such that <math>Q(t)=t</math>.<br />
<br />
==Problem 6==<br />
Let <math>P</math> be a convex <math>n</math>-sided polygon with vertices <math>V_1, V_2, \dots, V_n,</math> and sides <math>S_1, S_2, \dots, S_n.</math> For a given side <math>S_i,</math> let <math>A_i</math> be the maximum possible area of a triangle with vertices among <math>V_1, V_2, \dots, V_n</math> and with <math>S_i</math> as a side. Show that the sum of the areas <math>A_1, A_2, \dots, A_n</math> is at least twice the area of <math>P.</math><br />
<br />
==See Also==<br />
* [[2006 IMO]]<br />
* [[IMO Problems and Solutions]]<br />
* [[IMO]]<br />
<br />
{{IMO box|year=2006|before=[[2005 IMO Problems]]|after=[[2007 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2006_IMO_Problems&diff=1514122006 IMO Problems2021-04-13T01:28:20Z<p>Negia: /* Problem 2 */</p>
<hr />
<div>==Problem 1==<br />
Let <math>ABC</math> be a triangle with incentre <math>I.</math> A point <math>P</math> in the interior of the triangle satisfies <math>\angle PBA + \angle PCA = \angle PBC + \angle PCB</math>.<br />
Show that <math>AP \ge AI,</math> and that equality holds if and only if <math>P = I.</math><br />
<br />
==Problem 2==<br />
Let <math>P</math> be a regular 2006 sided polygon. A diagonal of <math>P</math> is called good if its endpoints divide the boundary of <math>P</math> into two parts, each composed of an odd number of sides of <math>P</math>. The sides of <math>P</math> are also called good. Suppose <math>P</math> has been dissected into triangles by 2003 diagonals, no two of which have a common point in the interior of <math>P</math>. Find the maximum number of isosceles triangles having two good sides that could appear in such a configuration.<br />
<br />
==Problem 3==<br />
Determine the least real number <math>M</math> such that the inequality <cmath> \left| ab\left(a^{2}-b^{2}\right)+bc\left(b^{2}-c^{2}\right)+ca\left(c^{2}-a^{2}\right)\right|\leq M\left(a^{2}+b^{2}+c^{2}\right)^{2} </cmath> holds for all real numbers <math>a,b</math> and <math>c</math><br />
<br />
==Problem 4==<br />
Determine all pairs <math>(x, y)</math> of integers such that <cmath>1+2^{x}+2^{2x+1}= y^{2}.</cmath><br />
<br />
==Problem 5==<br />
Let <math>P(x)</math> be a polynomial of degree <math>n>1</math> with integer coefficients, and let <math>k</math> be a positive integer. Consider the polynomial <math>Q(x) = P( P ( \ldots P(P(x)) \ldots ))</math>, where <math>P</math> occurs <math>k</math> times. Prove that there are at most <math>n</math> integers <math>t</math> such that <math>Q(t)=t</math>.<br />
<br />
==Problem 6==<br />
Assign to each side b of a convex polygon P the maximum area of a triangle that has b as a side and is contained in P. Show that the sum of the areas assigned to the sides of P is at least twice the area of P.<br />
<br />
==See Also==<br />
* [[2006 IMO]]<br />
* [[IMO Problems and Solutions]]<br />
* [[IMO]]<br />
<br />
{{IMO box|year=2006|before=[[2005 IMO Problems]]|after=[[2007 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2012_IMO_Problems&diff=1478552012 IMO Problems2021-02-24T15:00:46Z<p>Negia: /* Problem 1. */</p>
<hr />
<div>Problems of the 53st [[IMO]] 2012 in Mar del Plata, Argentina. <br />
<br />
== Day 1 ==<br />
=== Problem 1. ===<br />
Given <math>\triangle ABC</math> the point <math>J</math> is the centre of the excircle opposite the vertex <math>A</math>. This excircle is tangent to the side <math>BC</math> at <math>M</math>, and to the lines <math>AB</math> and <math>AC</math> at <math>K</math> and <math>L</math>, respectively. The lines <math>LM</math> and <math>BJ</math> meet at <math>F</math>, and the lines <math>KM</math> and <math>CJ</math> meet at <math>G</math>. Let <math>S</math> be the point of intersection of the lines <math>AF</math> and <math>BC</math>, and let <math>T</math> be the point of intersection of the lines <math>AG</math> and <math>BC</math>. Prove that <math>M</math> is the midpoint of <math>ST</math>.<br />
(The excircle of <math>ABC</math> opposite the vertex <math>A</math> is the circle that is tangent to the line segment <math>BC</math>,<br />
to the ray <math>AB</math> beyond <math>B</math>, and to the ray <math>AC</math> beyond <math>C</math>.)<br />
<br />
''Author: Evangelos Psychas, Greece''<br />
<br />
[[2012 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2. === <br />
Let <math>{{a}_{2}}, {{a}_{3}}, \cdots, {{a}_{n}}</math> be positive real numbers that satisfy <math>{{a}_{2}}\cdot {{a}_{3}}\cdots {{a}_{n}}=1</math> . Prove that<br />
<cmath> \left(a_2+1\right)^2\cdot \left(a_3+1\right)^3\cdots \left(a_n+1\right)^n\gneq n^n</cmath><br />
<br />
''Author: Angelo di Pasquale, Australia''<br />
<br />
[[2012 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3. ===<br />
The ''liar’s guessing game'' is a game played between two players <math>A</math> and <math>B</math>. The rules of the game depend on two positive integers <math>k</math> and <math>n</math> which are known to both players. At the start of the game A chooses integers <math>x</math> and <math>N</math> with <math>1\le x\le N</math>. Player <math>A</math> keeps <math>x</math> secret, and truthfully tells <math>N</math> to player <math>B</math>. Player <math>B</math> now tries to obtain information about <math>x</math> by asking player <math>A</math> questions as follows: each question consists of <math>B</math> specifying an arbitrary set <math>S</math> of positive integers (possibly one specified in some previous question), and asking <math>A</math> whether <math>x</math> belongs to <math>S</math>. Player <math>B</math> may ask as many such questions as he wishes. After each question, player <math>A</math> must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is<br />
that, among any <math>k + 1</math> consecutive answers, at least one answer must be truthful. After <math>B</math> has asked as many questions as he wants, he must specify a set <math>X</math> of at most <math>n</math> positive integers. If <math>x</math> belongs to <math>X</math>, then <math>B</math> wins; otherwise, he loses. Prove that:<br />
# If <math>n\ge {{2}^{k}}</math>, then <math>B</math> can guarantee a win.<br />
# For all sufficiently large <math>k</math>, there exists an integer <math>n\ge {1.99^k}</math> such that <math>B</math> cannot guarantee a win.<br />
<br />
''Author: David Arthur, Canada ''<br />
<br />
[[2012 IMO Problems/Problem 3 | Solution]]<br />
<br />
<br />
== Day 2 ==<br />
=== Problem 4. ===<br />
Find all functions <math>f:\mathbb{Z}\to \mathbb{Z}</math> such that, for all integers <math>a</math>, <math>b</math>, <math>c</math> that satisfy <math>a+b+c = 0</math>, the following equality holds:<br />
<cmath>f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).</cmath><br />
(Here <math>\mathbb{Z}</math> denotes the set of integers.)<br />
<br />
''Author: Liam Baker, South Africa ''<br />
<br />
[[2012 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5. ===<br />
Let <math>ABC</math> be a triangle with <math>\angle BCA=90{}^\circ </math>, and let <math>D</math> be the foot of the altitude from <math>C</math>. Let <math>X</math> be a point in the interior of the segment <math>CD</math>. Let K be the point on the segment <math>AX</math> such that <math>BK = BC</math>. Similarly, let <math>L</math> be the point on the segment <math>BX</math> such that <math>AL = AC</math>. Let <math>M</math> be the point of intersection of <math>AL</math> and <math>BK</math>.<br />
Show that <math>MK = ML</math>.<br />
<br />
''Author: Josef Tkadlec, Czech Republic''<br />
<br />
[[2012 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6. ===<br />
Find all positive integers n for which there exist non-negative integers <math>a_1</math>, <math>a_2</math>, <math>\ldots</math> , <math>a_n</math> such that<br />
<math>\frac{1}{{{2}^{{{a}_{1}}}}}+\frac{1}{{{2}^{{{a}_{2}}}}}+\cdots +\frac{1}{{{2}^{{{a}_{n}}}}}=\frac{1}{{{3}^{{{a}_{1}}}}}+\frac{2}{{{3}^{{{a}_{2}}}}}+\cdots +\frac{n}{{{3}^{{{a}_{n}}}}}=1</math><br />
<br />
''Author: Dušan Djukić, Serbia''<br />
<br />
[[2012 IMO Problems/Problem 6 | Solution]]<br />
<br />
== Resources ==<br />
* [[2012 IMO]]<br />
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=2012&sid=0b159fdf0134b865758167cc5cf255dd 2012 IMO Problems on the Resources page]<br />
<br />
{{IMO box|year=2012|before=[[2011 IMO Problems]]|after=[[2013 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2012_IMO_Problems&diff=1478542012 IMO Problems2021-02-24T15:00:23Z<p>Negia: /* Problem 1. */</p>
<hr />
<div>Problems of the 53st [[IMO]] 2012 in Mar del Plata, Argentina. <br />
<br />
== Day 1 ==<br />
=== Problem 1. ===<br />
Given triangle <math>ABC</math> the point <math>J</math> is the centre of the excircle opposite the vertex <math>A</math>. This excircle is tangent to the side <math>BC</math> at <math>M</math>, and to the lines <math>AB</math> and <math>AC</math> at <math>K</math> and <math>L</math>, respectively. The lines <math>LM</math> and <math>BJ</math> meet at <math>F</math>, and the lines <math>KM</math> and <math>CJ</math> meet at <math>G</math>. Let <math>S</math> be the point of intersection of the lines <math>AF</math> and <math>BC</math>, and let <math>T</math> be the point of intersection of the lines <math>AG</math> and <math>BC</math>. Prove that <math>M</math> is the midpoint of <math>ST</math>.<br />
(The excircle of <math>ABC</math> opposite the vertex <math>A</math> is the circle that is tangent to the line segment <math>BC</math>,<br />
to the ray <math>AB</math> beyond <math>B</math>, and to the ray <math>AC</math> beyond <math>C</math>.)<br />
<br />
''Author: Evangelos Psychas, Greece''<br />
<br />
[[2012 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2. === <br />
Let <math>{{a}_{2}}, {{a}_{3}}, \cdots, {{a}_{n}}</math> be positive real numbers that satisfy <math>{{a}_{2}}\cdot {{a}_{3}}\cdots {{a}_{n}}=1</math> . Prove that<br />
<cmath> \left(a_2+1\right)^2\cdot \left(a_3+1\right)^3\cdots \left(a_n+1\right)^n\gneq n^n</cmath><br />
<br />
''Author: Angelo di Pasquale, Australia''<br />
<br />
[[2012 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3. ===<br />
The ''liar’s guessing game'' is a game played between two players <math>A</math> and <math>B</math>. The rules of the game depend on two positive integers <math>k</math> and <math>n</math> which are known to both players. At the start of the game A chooses integers <math>x</math> and <math>N</math> with <math>1\le x\le N</math>. Player <math>A</math> keeps <math>x</math> secret, and truthfully tells <math>N</math> to player <math>B</math>. Player <math>B</math> now tries to obtain information about <math>x</math> by asking player <math>A</math> questions as follows: each question consists of <math>B</math> specifying an arbitrary set <math>S</math> of positive integers (possibly one specified in some previous question), and asking <math>A</math> whether <math>x</math> belongs to <math>S</math>. Player <math>B</math> may ask as many such questions as he wishes. After each question, player <math>A</math> must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is<br />
that, among any <math>k + 1</math> consecutive answers, at least one answer must be truthful. After <math>B</math> has asked as many questions as he wants, he must specify a set <math>X</math> of at most <math>n</math> positive integers. If <math>x</math> belongs to <math>X</math>, then <math>B</math> wins; otherwise, he loses. Prove that:<br />
# If <math>n\ge {{2}^{k}}</math>, then <math>B</math> can guarantee a win.<br />
# For all sufficiently large <math>k</math>, there exists an integer <math>n\ge {1.99^k}</math> such that <math>B</math> cannot guarantee a win.<br />
<br />
''Author: David Arthur, Canada ''<br />
<br />
[[2012 IMO Problems/Problem 3 | Solution]]<br />
<br />
<br />
== Day 2 ==<br />
=== Problem 4. ===<br />
Find all functions <math>f:\mathbb{Z}\to \mathbb{Z}</math> such that, for all integers <math>a</math>, <math>b</math>, <math>c</math> that satisfy <math>a+b+c = 0</math>, the following equality holds:<br />
<cmath>f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).</cmath><br />
(Here <math>\mathbb{Z}</math> denotes the set of integers.)<br />
<br />
''Author: Liam Baker, South Africa ''<br />
<br />
[[2012 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5. ===<br />
Let <math>ABC</math> be a triangle with <math>\angle BCA=90{}^\circ </math>, and let <math>D</math> be the foot of the altitude from <math>C</math>. Let <math>X</math> be a point in the interior of the segment <math>CD</math>. Let K be the point on the segment <math>AX</math> such that <math>BK = BC</math>. Similarly, let <math>L</math> be the point on the segment <math>BX</math> such that <math>AL = AC</math>. Let <math>M</math> be the point of intersection of <math>AL</math> and <math>BK</math>.<br />
Show that <math>MK = ML</math>.<br />
<br />
''Author: Josef Tkadlec, Czech Republic''<br />
<br />
[[2012 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6. ===<br />
Find all positive integers n for which there exist non-negative integers <math>a_1</math>, <math>a_2</math>, <math>\ldots</math> , <math>a_n</math> such that<br />
<math>\frac{1}{{{2}^{{{a}_{1}}}}}+\frac{1}{{{2}^{{{a}_{2}}}}}+\cdots +\frac{1}{{{2}^{{{a}_{n}}}}}=\frac{1}{{{3}^{{{a}_{1}}}}}+\frac{2}{{{3}^{{{a}_{2}}}}}+\cdots +\frac{n}{{{3}^{{{a}_{n}}}}}=1</math><br />
<br />
''Author: Dušan Djukić, Serbia''<br />
<br />
[[2012 IMO Problems/Problem 6 | Solution]]<br />
<br />
== Resources ==<br />
* [[2012 IMO]]<br />
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=2012&sid=0b159fdf0134b865758167cc5cf255dd 2012 IMO Problems on the Resources page]<br />
<br />
{{IMO box|year=2012|before=[[2011 IMO Problems]]|after=[[2013 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2012_IMO_Problems&diff=1478532012 IMO Problems2021-02-24T14:59:22Z<p>Negia: /* Problem 6. */</p>
<hr />
<div>Problems of the 53st [[IMO]] 2012 in Mar del Plata, Argentina. <br />
<br />
== Day 1 ==<br />
=== Problem 1. ===<br />
Given triangle ABC the point <math>J</math> is the centre of the excircle opposite the vertex <math>A</math>.<br />
This excircle is tangent to the side <math>BC</math> at <math>M</math>, and to the lines <math>AB</math> and <math>AC</math> at <math>K</math> and <math>L</math>, respectively.<br />
The lines <math>LM</math> and <math>BJ</math> meet at <math>F</math>, and the lines <math>KM</math> and <math>CJ</math> meet at <math>G</math>. Let <math>S</math> be the point of<br />
intersection of the lines <math>AF</math> and <math>BC</math>, and let <math>T</math> be the point of intersection of the lines <math>AG</math> and <math>BC</math>.<br />
Prove that <math>M</math> is the midpoint of <math>ST</math>.<br />
(The excircle of <math>ABC</math> opposite the vertex <math>A</math> is the circle that is tangent to the line segment <math>BC</math>,<br />
to the ray <math>AB</math> beyond <math>B</math>, and to the ray <math>AC</math> beyond <math>C</math>.)<br />
<br />
''Author: Evangelos Psychas, Greece''<br />
<br />
[[2012 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2. === <br />
Let <math>{{a}_{2}}, {{a}_{3}}, \cdots, {{a}_{n}}</math> be positive real numbers that satisfy <math>{{a}_{2}}\cdot {{a}_{3}}\cdots {{a}_{n}}=1</math> . Prove that<br />
<cmath> \left(a_2+1\right)^2\cdot \left(a_3+1\right)^3\cdots \left(a_n+1\right)^n\gneq n^n</cmath><br />
<br />
''Author: Angelo di Pasquale, Australia''<br />
<br />
[[2012 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3. ===<br />
The ''liar’s guessing game'' is a game played between two players <math>A</math> and <math>B</math>. The rules of the game depend on two positive integers <math>k</math> and <math>n</math> which are known to both players. At the start of the game A chooses integers <math>x</math> and <math>N</math> with <math>1\le x\le N</math>. Player <math>A</math> keeps <math>x</math> secret, and truthfully tells <math>N</math> to player <math>B</math>. Player <math>B</math> now tries to obtain information about <math>x</math> by asking player <math>A</math> questions as follows: each question consists of <math>B</math> specifying an arbitrary set <math>S</math> of positive integers (possibly one specified in some previous question), and asking <math>A</math> whether <math>x</math> belongs to <math>S</math>. Player <math>B</math> may ask as many such questions as he wishes. After each question, player <math>A</math> must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is<br />
that, among any <math>k + 1</math> consecutive answers, at least one answer must be truthful. After <math>B</math> has asked as many questions as he wants, he must specify a set <math>X</math> of at most <math>n</math> positive integers. If <math>x</math> belongs to <math>X</math>, then <math>B</math> wins; otherwise, he loses. Prove that:<br />
# If <math>n\ge {{2}^{k}}</math>, then <math>B</math> can guarantee a win.<br />
# For all sufficiently large <math>k</math>, there exists an integer <math>n\ge {1.99^k}</math> such that <math>B</math> cannot guarantee a win.<br />
<br />
''Author: David Arthur, Canada ''<br />
<br />
[[2012 IMO Problems/Problem 3 | Solution]]<br />
<br />
<br />
== Day 2 ==<br />
=== Problem 4. ===<br />
Find all functions <math>f:\mathbb{Z}\to \mathbb{Z}</math> such that, for all integers <math>a</math>, <math>b</math>, <math>c</math> that satisfy <math>a+b+c = 0</math>, the following equality holds:<br />
<cmath>f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).</cmath><br />
(Here <math>\mathbb{Z}</math> denotes the set of integers.)<br />
<br />
''Author: Liam Baker, South Africa ''<br />
<br />
[[2012 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5. ===<br />
Let <math>ABC</math> be a triangle with <math>\angle BCA=90{}^\circ </math>, and let <math>D</math> be the foot of the altitude from <math>C</math>. Let <math>X</math> be a point in the interior of the segment <math>CD</math>. Let K be the point on the segment <math>AX</math> such that <math>BK = BC</math>. Similarly, let <math>L</math> be the point on the segment <math>BX</math> such that <math>AL = AC</math>. Let <math>M</math> be the point of intersection of <math>AL</math> and <math>BK</math>.<br />
Show that <math>MK = ML</math>.<br />
<br />
''Author: Josef Tkadlec, Czech Republic''<br />
<br />
[[2012 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6. ===<br />
Find all positive integers n for which there exist non-negative integers <math>a_1</math>, <math>a_2</math>, <math>\ldots</math> , <math>a_n</math> such that<br />
<math>\frac{1}{{{2}^{{{a}_{1}}}}}+\frac{1}{{{2}^{{{a}_{2}}}}}+\cdots +\frac{1}{{{2}^{{{a}_{n}}}}}=\frac{1}{{{3}^{{{a}_{1}}}}}+\frac{2}{{{3}^{{{a}_{2}}}}}+\cdots +\frac{n}{{{3}^{{{a}_{n}}}}}=1</math><br />
<br />
''Author: Dušan Djukić, Serbia''<br />
<br />
[[2012 IMO Problems/Problem 6 | Solution]]<br />
<br />
== Resources ==<br />
* [[2012 IMO]]<br />
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=2012&sid=0b159fdf0134b865758167cc5cf255dd 2012 IMO Problems on the Resources page]<br />
<br />
{{IMO box|year=2012|before=[[2011 IMO Problems]]|after=[[2013 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2004_IMO_Problems&diff=1476332004 IMO Problems2021-02-21T08:36:43Z<p>Negia: /* Problem 4 */</p>
<hr />
<div>Problems of the 45th [[IMO]] 2004 Athens, Greece.<br />
<br />
== Day 1 ==<br />
<br />
=== Problem 1 ===<br />
Let <math>ABC</math> be an acute-angled triangle with <math>AB\neq AC</math>. The circle with diameter <math>BC</math> intersects the sides <math>AB</math> and <math>AC</math> at <math>M</math> and <math>N</math> respectively. Denote by <math>O</math> the midpoint of the side <math>BC</math>. The bisectors of the angles <math>\angle BAC</math> and <math>\angle MON</math> intersect at <math>R</math>. Prove that the circumcircles of the triangles <math>BMR</math> and <math>CNR</math> have a common point lying on the side <math>BC</math>.<br />
<br />
[[2004 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2 ===<br />
Find all polynomials <math>f</math> with real coefficients such that for all reals <math>a,b,c</math> such that <math>ab + bc + ca = 0</math> we have the following relations<br />
<br />
<cmath>f(a - b) + f(b - c) + f(c - a) = 2f(a + b + c).</cmath><br />
<br />
[[2004 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3 ===<br />
Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.<br />
<br />
<asy><br />
unitsize(0.5 cm);<br />
<br />
draw((0,0)--(1,0));<br />
draw((0,1)--(1,1));<br />
draw((2,1)--(3,1));<br />
draw((0,2)--(3,2));<br />
draw((0,3)--(3,3));<br />
draw((0,0)--(0,3));<br />
draw((1,0)--(1,3));<br />
draw((2,1)--(2,3));<br />
draw((3,1)--(3,3));<br />
</asy><br />
<br />
Determine all <math>m \times n</math> rectangles that can be covered without gaps and without overlaps with hooks such that;<br />
(a) the rectangle is covered without gaps and without overlaps,<br />
(b) no part of a hook covers area outside the rectangle.<br />
<br />
[[2004 IMO Problems/Problem 3 | Solution]]<br />
<br />
== Day 2 ==<br />
<br />
=== Problem 4 ===<br />
Let <math>n \geq 3</math> be an integer. Let <math>t_1, t_2, \dots ,t_n</math> be positive real numbers such that<br />
<br />
<cmath>n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right).</cmath><br />
Show that <math>t_i</math>, <math>t_j</math>, <math>t_k</math> are side lengths of a triangle for all <math>i</math>, <math>j</math>, <math>k</math> with <math>1 \leq i < j < k \leq n</math>.<br />
<br />
[[2004 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5 ===<br />
In a convex quadrilateral <math>ABCD</math>, the diagonal <math>BD</math> bisects neither the angle <math>ABC</math> nor the angle <math>CDA</math>. The point <math>P</math> lies inside <math>ABCD</math> and satisfies<cmath>\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.</cmath><br />
Prove that <math>ABCD</math> is a cyclic quadrilateral if and only if <math>AP = CP.</math><br />
<br />
[[2004 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6 ===<br />
We call a positive integer alternating if every two consecutive digits in its decimal representation have a different parity.<br />
<br />
Find all positive integers <math>n</math> such that <math>n</math> has a multiple which is alternating.<br />
<br />
[[2004 IMO Problems/Problem 6 | Solution]]<br />
<br />
== Resources ==<br />
<br />
* [[2004 IMO]]<br />
* [https://artofproblemsolving.com/community/c3831_2004_imo IMO 2004 Problems on the Resources page]<br />
<br />
{{IMO box|year=2004|before=[[2003 IMO Problems]]|after=[[2005 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2004_IMO_Problems&diff=1476322004 IMO Problems2021-02-21T08:36:24Z<p>Negia: /* Problem 4 */</p>
<hr />
<div>Problems of the 45th [[IMO]] 2004 Athens, Greece.<br />
<br />
== Day 1 ==<br />
<br />
=== Problem 1 ===<br />
Let <math>ABC</math> be an acute-angled triangle with <math>AB\neq AC</math>. The circle with diameter <math>BC</math> intersects the sides <math>AB</math> and <math>AC</math> at <math>M</math> and <math>N</math> respectively. Denote by <math>O</math> the midpoint of the side <math>BC</math>. The bisectors of the angles <math>\angle BAC</math> and <math>\angle MON</math> intersect at <math>R</math>. Prove that the circumcircles of the triangles <math>BMR</math> and <math>CNR</math> have a common point lying on the side <math>BC</math>.<br />
<br />
[[2004 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2 ===<br />
Find all polynomials <math>f</math> with real coefficients such that for all reals <math>a,b,c</math> such that <math>ab + bc + ca = 0</math> we have the following relations<br />
<br />
<cmath>f(a - b) + f(b - c) + f(c - a) = 2f(a + b + c).</cmath><br />
<br />
[[2004 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3 ===<br />
Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.<br />
<br />
<asy><br />
unitsize(0.5 cm);<br />
<br />
draw((0,0)--(1,0));<br />
draw((0,1)--(1,1));<br />
draw((2,1)--(3,1));<br />
draw((0,2)--(3,2));<br />
draw((0,3)--(3,3));<br />
draw((0,0)--(0,3));<br />
draw((1,0)--(1,3));<br />
draw((2,1)--(2,3));<br />
draw((3,1)--(3,3));<br />
</asy><br />
<br />
Determine all <math>m \times n</math> rectangles that can be covered without gaps and without overlaps with hooks such that;<br />
(a) the rectangle is covered without gaps and without overlaps,<br />
(b) no part of a hook covers area outside the rectangle.<br />
<br />
[[2004 IMO Problems/Problem 3 | Solution]]<br />
<br />
== Day 2 ==<br />
<br />
=== Problem 4 ===<br />
Let <math>n \geq 3</math> be an integer. Let <math>t_1, t_2, \cdots ,t_n</math> be positive real numbers such that<br />
<br />
<cmath>n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right).</cmath><br />
Show that <math>t_i</math>, <math>t_j</math>, <math>t_k</math> are side lengths of a triangle for all <math>i</math>, <math>j</math>, <math>k</math> with <math>1 \leq i < j < k \leq n</math>.<br />
<br />
[[2004 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5 ===<br />
In a convex quadrilateral <math>ABCD</math>, the diagonal <math>BD</math> bisects neither the angle <math>ABC</math> nor the angle <math>CDA</math>. The point <math>P</math> lies inside <math>ABCD</math> and satisfies<cmath>\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.</cmath><br />
Prove that <math>ABCD</math> is a cyclic quadrilateral if and only if <math>AP = CP.</math><br />
<br />
[[2004 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6 ===<br />
We call a positive integer alternating if every two consecutive digits in its decimal representation have a different parity.<br />
<br />
Find all positive integers <math>n</math> such that <math>n</math> has a multiple which is alternating.<br />
<br />
[[2004 IMO Problems/Problem 6 | Solution]]<br />
<br />
== Resources ==<br />
<br />
* [[2004 IMO]]<br />
* [https://artofproblemsolving.com/community/c3831_2004_imo IMO 2004 Problems on the Resources page]<br />
<br />
{{IMO box|year=2004|before=[[2003 IMO Problems]]|after=[[2005 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2004_IMO_Problems&diff=1476212004 IMO Problems2021-02-21T03:53:10Z<p>Negia: </p>
<hr />
<div>Problems of the 45th [[IMO]] 2004 Athens, Greece.<br />
<br />
== Day 1 ==<br />
<br />
=== Problem 1 ===<br />
Let <math>ABC</math> be an acute-angled triangle with <math>AB\neq AC</math>. The circle with diameter <math>BC</math> intersects the sides <math>AB</math> and <math>AC</math> at <math>M</math> and <math>N</math> respectively. Denote by <math>O</math> the midpoint of the side <math>BC</math>. The bisectors of the angles <math>\angle BAC</math> and <math>\angle MON</math> intersect at <math>R</math>. Prove that the circumcircles of the triangles <math>BMR</math> and <math>CNR</math> have a common point lying on the side <math>BC</math>.<br />
<br />
[[2004 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2 ===<br />
Find all polynomials <math>f</math> with real coefficients such that for all reals <math>a,b,c</math> such that <math>ab + bc + ca = 0</math> we have the following relations<br />
<br />
<cmath>f(a - b) + f(b - c) + f(c - a) = 2f(a + b + c).</cmath><br />
<br />
[[2004 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3 ===<br />
Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.<br />
<br />
<asy><br />
unitsize(0.5 cm);<br />
<br />
draw((0,0)--(1,0));<br />
draw((0,1)--(1,1));<br />
draw((2,1)--(3,1));<br />
draw((0,2)--(3,2));<br />
draw((0,3)--(3,3));<br />
draw((0,0)--(0,3));<br />
draw((1,0)--(1,3));<br />
draw((2,1)--(2,3));<br />
draw((3,1)--(3,3));<br />
</asy><br />
<br />
Determine all <math>m \times n</math> rectangles that can be covered without gaps and without overlaps with hooks such that;<br />
(a) the rectangle is covered without gaps and without overlaps,<br />
(b) no part of a hook covers area outside the rectangle.<br />
<br />
[[2004 IMO Problems/Problem 3 | Solution]]<br />
<br />
== Day 2 ==<br />
<br />
=== Problem 4 ===<br />
Let <math>n \geq 3</math> be an integer. Let <math>t_1</math>, <math>t_2</math>, ..., <math>t_n</math> be positive real numbers such that<br />
<br />
<cmath>n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right).</cmath><br />
Show that <math>t_i</math>, <math>t_j</math>, <math>t_k</math> are side lengths of a triangle for all <math>i</math>, <math>j</math>, <math>k</math> with <math>1 \leq i < j < k \leq n</math>.<br />
<br />
[[2004 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5 ===<br />
In a convex quadrilateral <math>ABCD</math>, the diagonal <math>BD</math> bisects neither the angle <math>ABC</math> nor the angle <math>CDA</math>. The point <math>P</math> lies inside <math>ABCD</math> and satisfies<cmath>\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.</cmath><br />
Prove that <math>ABCD</math> is a cyclic quadrilateral if and only if <math>AP = CP.</math><br />
<br />
[[2004 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6 ===<br />
We call a positive integer alternating if every two consecutive digits in its decimal representation have a different parity.<br />
<br />
Find all positive integers <math>n</math> such that <math>n</math> has a multiple which is alternating.<br />
<br />
[[2004 IMO Problems/Problem 6 | Solution]]<br />
<br />
== Resources ==<br />
<br />
* [[2004 IMO]]<br />
* [https://artofproblemsolving.com/community/c3831_2004_imo IMO 2004 Problems on the Resources page]<br />
<br />
{{IMO box|year=2004|before=[[2003 IMO Problems]]|after=[[2005 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2004_IMO_Problems/Problem_4&diff=1476202004 IMO Problems/Problem 42021-02-21T03:50:43Z<p>Negia: </p>
<hr />
<div>== Problem ==<br />
(''Hojoo Lee'') Let <math>n \geq 3</math> be an integer. Let <math>t_1, t_2, \dots , t_n</math> be positive real numbers such that <br />
<br />
<cmath>n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right).</cmath><br />
<br />
Show that <math>t_i</math>, <math>t_j</math>, <math>t_k</math> are side lengths of a triangle for all <math>i</math>, <math>j</math>, <math>k</math> with <math>1 \leq i < j < k \leq n</math>.<br />
<br />
== Solution ==<br />
For <math>n=3</math>, suppose (for sake of contradiction) that <math>t_3 = t_2 + t_1 + k</math> for <math>k \ge 0</math>; then (by [[Cauchy-Schwarz Inequality]])<br />
<br />
<cmath>\begin{align*}10 &> [2(t_1 + t_2) + k]\left(\frac {1}{t_1} + \frac {1}{t_2} + \frac 1{t_1 + t_2 + k}\right) = 2(t_1+t_2)\left(\frac 1{t_1} + \frac{1}{t_2}\right) + \left(\frac{k}{t_1} + \frac k{t_2}\right) + \frac{2t_1 + 2t_2 + k}{t_1 + t_2 + k}\\ &\ge 8 + \left(\frac{k}{t_1} + \frac k{t_2}\right) + 2 - \frac{k}{t_1 + t_2 + k}\\ &= 10 + k\left(\frac{1}{t_1} + \frac {1}{t_2} - \frac{1}{t_1 + t_2+k}\right) \ge 10\end{align*}</cmath><br />
<br />
so it is true for <math>n=3</math>. We now claim the result by induction; for <math>n \ge 4</math>, we have <br />
<br />
<cmath>\begin{align*}f(n) &:= \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right) \\ &= \left( t_1 + t_2 + ... + t_{n-1} \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_{n-1}} \right) + t_n \sum_{i=1}^{n-1} \frac 1{t_i} + \frac{1}{t_n} \sum_{i=1}^{n-1} t_i + 1\end{align*}</cmath><br />
<br />
By [[AM-GM]], <math>\frac{t_n}{t_i} + \frac{t_i}{t_n} \ge 2</math>, so <math>f(n) \ge f(n-1) + 2(n-1) + 1 = f(n-1) + 2n - 1</math>. Then the problem is reduced to proving the statement true for <math>n-1</math> numbers, as desired.<math>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\square</math><br />
<br />
== See also ==<br />
*<url>viewtopic.php?p=99756#99756 AoPS/MathLinks discussion</url><br />
<br />
[[Category:Olympiad Algebra Problems]]</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2004_IMO_Problems/Problem_3&diff=1476182004 IMO Problems/Problem 32021-02-21T03:45:06Z<p>Negia: </p>
<hr />
<div>== Problem ==<br />
Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.<br />
<br />
<asy><br />
unitsize(0.5 cm);<br />
<br />
draw((0,0)--(1,0));<br />
draw((0,1)--(1,1));<br />
draw((2,1)--(3,1));<br />
draw((0,2)--(3,2));<br />
draw((0,3)--(3,3));<br />
draw((0,0)--(0,3));<br />
draw((1,0)--(1,3));<br />
draw((2,1)--(2,3));<br />
draw((3,1)--(3,3));<br />
</asy><br />
<br />
Determine all <math>m \times n</math> rectangles that can be covered without gaps and without overlaps with hooks such that;<br />
(a) the rectangle is covered without gaps and without overlaps,<br />
(b) no part of a hook covers area outside the rectangle.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2004_IMO_Problems/Problem_3&diff=1476162004 IMO Problems/Problem 32021-02-21T03:32:03Z<p>Negia: </p>
<hr />
<div>Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.<br />
<br />
<asy><br />
unitsize(0.5 cm);<br />
<br />
draw((0,0)--(1,0));<br />
draw((0,1)--(1,1));<br />
draw((2,1)--(3,1));<br />
draw((0,2)--(3,2));<br />
draw((0,3)--(3,3));<br />
draw((0,0)--(0,3));<br />
draw((1,0)--(1,3));<br />
draw((2,1)--(2,3));<br />
draw((3,1)--(3,3));<br />
</asy><br />
<br />
Determine all <math>m \times n</math> rectangles that can be covered without gaps and without overlaps with hooks such that;<br />
(a) the rectangle is covered without gaps and without overlaps,<br />
(b) no part of a hook covers area outside the rectangle.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2004_IMO_Problems/Problem_3&diff=1476142004 IMO Problems/Problem 32021-02-21T03:29:41Z<p>Negia: </p>
<hr />
<div>Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.<br />
<br />
[asy]<br />
unitsize(0.5 cm);<br />
<br />
draw((0,0)--(1,0));<br />
draw((0,1)--(1,1));<br />
draw((2,1)--(3,1));<br />
draw((0,2)--(3,2));<br />
draw((0,3)--(3,3));<br />
draw((0,0)--(0,3));<br />
draw((1,0)--(1,3));<br />
draw((2,1)--(2,3));<br />
draw((3,1)--(3,3));<br />
[/asy]<br />
Determine all <math>m \times n</math> rectangles that can be covered without gaps and without overlaps with hooks such that;<br />
(a) the rectangle is covered without gaps and without overlaps,<br />
(b) no part of a hook covers area outside the rectangle.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2004_IMO_Problems/Problem_3&diff=1476132004 IMO Problems/Problem 32021-02-21T03:28:06Z<p>Negia: </p>
<hr />
<div>Define a “hook” to be a figure made up of six unit squares as shown in this picture: http://bit.ly/IMO2004P1, or any of the figures obtained by applying rotations and reflections to this figure.<br />
<br />
Which <math>m\times n</math> rectangles can be tiled by hooks?</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2004_IMO_Problems&diff=1476112004 IMO Problems2021-02-21T03:25:48Z<p>Negia: Created page with "Problems of the 46th IMO 2004 Mérida, Mexico. == Day 1 == === Problem 1 === Six points are chosen on the sides of an equilateral triangle <math>ABC</math>: <math>A_1, A..."</p>
<hr />
<div>Problems of the 46th [[IMO]] 2004 Mérida, Mexico.<br />
<br />
== Day 1 ==<br />
<br />
=== Problem 1 ===<br />
Six points are chosen on the sides of an equilateral triangle <math>ABC</math>: <math>A_1, A_2</math> on <math>BC</math>, <math>B_1</math>, <math>B_2</math> on <math>CA</math> and <math>C_1</math>, <math>C_2</math> on <math>AB</math>, such that they are the vertices of a convex hexagon <math>A_1A_2B_1B_2C_1C_2</math> with equal side lengths. Prove that the lines <math>A_1B_2, B_1C_2</math> and <math>C_1A_2</math> are concurrent.<br />
<br />
[[2004 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2 ===<br />
Let <math>a_1, a_2, \dots</math> be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer <math>n</math> the numbers <math>a_1, a_2, \dots, a_n</math> leave <math>n</math> different remainders upon division by <math>n</math>. Prove that every integer occurs exactly once in the sequence.<br />
<br />
[[2004 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3 ===<br />
Let <math>x, y, z > 0</math> satisfy <math>xyz\ge 1</math>. Prove that<cmath>\frac{x^5-x^2}{x^5+y^2+z^2} + \frac{y^5-y^2}{x^2+y^5+z^2} + \frac{z^5-z^2}{x^2+y^2+z^5} \ge 0.</cmath><br />
<br />
[[2004 IMO Problems/Problem 3 | Solution]]<br />
<br />
== Day 2 ==<br />
<br />
=== Problem 4 ===<br />
Determine all positive integers relatively prime to all the terms of the infinite sequence<cmath>a_n=2^n+3^n+6^n -1,\ n\geq 1.</cmath><br />
<br />
[[2004 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5 ===<br />
Let <math>ABCD</math> be a fixed convex quadrilateral with <math>BC = DA</math> and <math>BC \nparallel DA</math>. Let two variable points <math>E</math> and <math>F</math> lie of the sides <math>BC</math> and <math>DA</math>, respectively, and satisfy <math>BE = DF</math>. The lines <math>AC</math> and <math>BD</math> meet at <math>P</math>, the lines <math>BD</math> and <math>EF</math> meet at <math>Q</math>, the lines <math>EF</math> and <math>AC</math> meet at <math>R</math>. Prove that the circumcircles of the triangles <math>PQR</math>, as <math>E</math> and <math>F</math> vary, have a common point other than <math>P</math>.<br />
<br />
[[2004 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6 ===<br />
In a mathematical competition, in which 6 problems were posed to the participants, every two of these problems were solved by more than 2/5 of the contestants. Moreover, no contestant solved all the 6 problems. Show that there are at least 2 contestants who solved exactly 5 problems each.<br />
<br />
[[2004 IMO Problems/Problem 6 | Solution]]<br />
<br />
== Resources ==<br />
<br />
* [[2004 IMO]]<br />
* [https://artofproblemsolving.com/community/c3831_2004_imo IMO 2004 Problems on the Resources page]<br />
<br />
{{IMO box|year=2004|before=[[2003 IMO Problems]]|after=[[2005 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=User:Piphi&diff=147608User:Piphi2021-02-21T03:15:02Z<p>Negia: /* User Count */</p>
<hr />
<div>{{User:Piphi/Template:Header}}<br />
<br><br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">464</font></center><br />
</div><br />
<div style="border:2px solid black; background:#919293;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black"><font color="black"><br />
Piphi is the creator of the [[User:Piphi/Games|AoPS Wiki Games by Piphi]], the future of games on AoPS.<br><br />
<br />
Piphi started the signature trend at around May 2020.<br><br />
<br />
Piphi has been very close to winning multiple [[Greed Control]] games, piphi placed 5th in game #18 and 2nd in game #19. Thanks to piphi, Greed Control games have started to be kept track of. Piphi made a spreadsheet that has all of Greed Control history [https://artofproblemsolving.com/community/c19451h2126208p15569802 here].<br><br />
<br />
Piphi also found out who won [[Reaper]] games #1 and #2 as seen [https://artofproblemsolving.com/community/c19451h1826745p15526330 here].<br><br />
<br />
Piphi created the [[AoPS Administrators]] page, added most of the AoPS Admins to it, and created the scrollable table.<br><br />
<br />
Piphi has also added a lot of the info that is in the [[Reaper Archives]].<br><br />
<br />
Piphi has a side-project that is making the Wiki's [[Main Page]] look better, you can check that out [[User:Piphi/AoPS Wiki|here]].<br><br />
<br />
Piphi published Greed Control Game 19 statistics [https://artofproblemsolving.com/community/c19451h2126212 here].<br />
<br />
Piphi has a post that was made an announcement in a official AoPS Forum [https://artofproblemsolving.com/community/c68h2175116 here].<br />
<br />
Piphi is a proud member of [https://artofproblemsolving.com/community/c562043 The Interuniversal GMAAS Society].<br />
<br />
Piphi has won 2 gold trophies at the [https://artofproblemsolving.com/community/c1124279 Asymptote Competition] and is now part of the staff.<br />
</font></div><br />
</div><br />
<div style="border:2px solid black; background:#333333;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="#f0f2f3" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
<div style="margin-left: 10px; margin-right: 10px; margin-bottom:10px"><font color="#f0f2f3"><br />
You can check out more goals/statistics [[User:Piphi/Statistics|here]].<br />
<br />
A User Count of 500<br />
{{User:Piphi/Template:Progress_Bar|91.6|width=100%}}<br />
<br />
200 subpages of [[User:Piphi]]<br />
{{User:Piphi/Template:Progress_Bar|97|width=100%}}<br />
<br />
200 signups for [[User:Piphi/Games|AoPS Wiki Games by Piphi]]<br />
{{User:Piphi/Template:Progress_Bar|50.5|width=100%}}<br />
<br />
Make 10,000 edits<br />
{{User:Piphi/Template:Progress_Bar|21.65|width=100%}}</font></div><br />
</div></div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2005_IMO_Problems&diff=1474382005 IMO Problems2021-02-19T06:08:37Z<p>Negia: Created page with "Problems of the 46th IMO 2005 Mérida, Mexico. == Day 1 == === Problem 1 === Six points are chosen on the sides of an equilateral triangle <math>ABC</math>: <math>A_1, A..."</p>
<hr />
<div>Problems of the 46th [[IMO]] 2005 Mérida, Mexico.<br />
<br />
== Day 1 ==<br />
<br />
=== Problem 1 ===<br />
Six points are chosen on the sides of an equilateral triangle <math>ABC</math>: <math>A_1, A_2</math> on <math>BC</math>, <math>B_1</math>, <math>B_2</math> on <math>CA</math> and <math>C_1</math>, <math>C_2</math> on <math>AB</math>, such that they are the vertices of a convex hexagon <math>A_1A_2B_1B_2C_1C_2</math> with equal side lengths. Prove that the lines <math>A_1B_2, B_1C_2</math> and <math>C_1A_2</math> are concurrent.<br />
<br />
[[2005 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2 ===<br />
Let <math>a_1, a_2, \dots</math> be a sequence of integers with infinitely many positive and negative terms. Suppose that for every positive integer <math>n</math> the numbers <math>a_1, a_2, \dots, a_n</math> leave <math>n</math> different remainders upon division by <math>n</math>. Prove that every integer occurs exactly once in the sequence.<br />
<br />
[[2005 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3 ===<br />
Let <math>x, y, z > 0</math> satisfy <math>xyz\ge 1</math>. Prove that<cmath>\frac{x^5-x^2}{x^5+y^2+z^2} + \frac{y^5-y^2}{x^2+y^5+z^2} + \frac{z^5-z^2}{x^2+y^2+z^5} \ge 0.</cmath><br />
<br />
[[2005 IMO Problems/Problem 3 | Solution]]<br />
<br />
== Day 2 ==<br />
<br />
=== Problem 4 ===<br />
Determine all positive integers relatively prime to all the terms of the infinite sequence<cmath>a_n=2^n+3^n+6^n -1,\ n\geq 1.</cmath><br />
<br />
[[2005 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5 ===<br />
Let <math>ABCD</math> be a fixed convex quadrilateral with <math>BC = DA</math> and <math>BC \nparallel DA</math>. Let two variable points <math>E</math> and <math>F</math> lie of the sides <math>BC</math> and <math>DA</math>, respectively, and satisfy <math>BE = DF</math>. The lines <math>AC</math> and <math>BD</math> meet at <math>P</math>, the lines <math>BD</math> and <math>EF</math> meet at <math>Q</math>, the lines <math>EF</math> and <math>AC</math> meet at <math>R</math>. Prove that the circumcircles of the triangles <math>PQR</math>, as <math>E</math> and <math>F</math> vary, have a common point other than <math>P</math>.<br />
<br />
[[2005 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6 ===<br />
In a mathematical competition, in which 6 problems were posed to the participants, every two of these problems were solved by more than 2/5 of the contestants. Moreover, no contestant solved all the 6 problems. Show that there are at least 2 contestants who solved exactly 5 problems each.<br />
<br />
[[2005 IMO Problems/Problem 6 | Solution]]<br />
<br />
== Resources ==<br />
<br />
* [[2005 IMO]]<br />
* [https://artofproblemsolving.com/community/c3832_2005_imo IMO 2005 Problems on the Resources page]<br />
<br />
{{IMO box|year=2005|before=[[2004 IMO Problems]]|after=[[2006 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2008_IMO_Problems/Problem_3&diff=1473622008 IMO Problems/Problem 32021-02-18T10:51:30Z<p>Negia: /* Solution */</p>
<hr />
<div>== Problem ==<br />
Prove that there are infinitely many positive integers <math>n</math> such that <math>n^{2} + 1</math> has a prime divisor greater than <math>2n + \sqrt {2n}</math>.<br />
<br />
== Solutions ==<br />
<br />
'''Solution 1'''<br />
<br />
The main idea is to take a gaussian prime <math>a+bi</math> and multiply it by a "twice as small" <math>c+di</math> to get <math>n+i</math>. The rest is just making up the little details.<br />
<br />
For each ''sufficiently large'' prime <math>p</math> of the form <math>4k+1</math>, we shall find a corresponding <math>n</math> such that <math>p</math> divides <math>n^2+1</math> and <math>p>2n+\sqrt{2n}</math>. Since there exist infinitely many such primes and, for each of them, <math>n \ge \sqrt{p-1}</math>, we will have found infinitely many ''distinct'' <math>n</math> satisfying the hypothesis.<br />
<br />
Take a prime <math>p</math> of the form <math>4k+1</math> and consider its "sum-of-two squares" representation <math>p=a^2+b^2</math>, which we know to exist for all such primes. As <math>a\ne b</math>, assume without loss of generality that <math>b>a</math>. If <math>a=1</math>, then <math>n=b</math> is what we are looking for, and <math>p=n^2+1 > 2n+\sqrt{2n}</math> as long as <math>p</math> (and hence <math>n</math>) is large enough. Assume from now on that <math>b>a>1</math>.<br />
<br />
Since <math>a</math> and <math>b</math> are (obviously) co-prime, there must exist integers <math>c</math> and <math>d</math> such that<br />
<cmath>ad+bc=1. \quad(1)</cmath><br />
In fact, if <math>c</math> and <math>d</math> are such numbers, then <math>c\pm ma</math> and <math>d\mp mb</math> work as well for any integer <math>m</math>, so we can assume that <math>c \in \left[-\frac{a}{2}, \frac{a}{2}\right]</math>.<br />
<br />
Define <math>n=|ac-bd|</math> and let's see why this is a good choice. For starters, notice that <math>(a^2+b^2)(c^2+d^2)=n^2+1</math>.<br />
<br />
If <math>c=\pm\frac{a}{2}</math>, from (1) we see that <math>a</math> must divide <math>2</math> and hence <math>a=2</math>. This implies, <math>d=-\frac{b-1}{2}</math> and <math>n=\frac{b(b-1)}{2}-2</math>. Therefore, <br />
<math>\left(b-\frac{1}{2}\right)^2 = 1/4 + 2(n+2) > 2n</math>, so <math>b > \sqrt{2n}+\frac{1}{2}</math>. Finally, <math>p=b^2+2^2 > 2n+\sqrt{2n}</math> and the case <math>c=\pm\frac{a}{2}</math> is cleared.<br />
<br />
We can safely assume now that<br />
<cmath>|c| \le \frac{a-1}{2}.</cmath><br />
As <math>b>a>1</math> implies <math>b>2</math>, we have<br />
<cmath>|d| = \left|\frac{1-bc}{a}\right| \le \frac{b(a-1)+2}{2a} < \frac{ba}{2a} = \frac{b}{2},</cmath><br />
so<br />
<cmath>|d| \le \frac{b-1}{2}.</cmath><br />
<br />
Therefore,<br />
<cmath>n^2+1 = (a^2+b^2)(c^2+d^2) \le p\left( \frac{(a-1)^2}{4}+\frac{(b-1)^2}{4} \right). \quad (2)</cmath> <br />
<br />
Before we proceed, we would like to show first that <math>a+b-1 > \sqrt{p}</math>. Observe that the function <math>x+\sqrt{p-x^2}</math> over <math>x\in(2,\sqrt{p-4})</math> reaches its minima on the ends, so <math>a+b</math> given <math>a^2+b^2=p</math> is minimized for <math>a = 2</math>, where it equals <math>2+\sqrt{p-2^2}</math>. So we want to show that <cmath>2+\sqrt{p-4} > \sqrt{p} + 1,</cmath><br />
which is not hard to show for ''large enough'' <math>p</math>.<br />
<br />
Now armed with <math>a+b-1>\sqrt{p}</math> and (2), we get<br />
<cmath>4(n^2+1)\le p( a^2+b^2 - 2(a+b-1) )< p( p-2\sqrt{p} )< u^2(u-1)^2,</cmath><br />
where <math>u=\sqrt{p}.</math><br />
<br />
Finally, <br />
<cmath>u^2(u-1)^2 > 4n^2+4 > 4n^2\Rightarrow<br />
u(u-1) > 2n \Rightarrow <br />
(u-\frac{1}{2})^2 > 2n+\frac{1}{4} > 2n \Rightarrow<br />
u > \sqrt{2n} + \frac{1}{2} \Rightarrow <br />
p = u^2 > 2n + \sqrt{2n}.</cmath><br />
The proof is complete.<br />
<br />
'''Solution 2'''<br />
<br />
We begin with a lemma.<br />
<br />
Lemma: For every prime <math>p \equiv 1 \mod{4}</math>, there exists a positive integer <math>n \le \dfrac{p - \sqrt{p-4}}{2}</math> such that <math>p\mid n^2 + 1</math>.<br />
<br />
Proof:<br />
We know that there must be a solution <math>x</math> to the equation<br />
<cmath>x^2+1 \equiv 0 \mod{p}</cmath><br />
with <math>1 \le x \le p-1</math>. However, if <math>x</math> is a solution then so is <math>p-x</math>, so there must be a solution <math>x \le p-1</math>. Let <math>n</math> denote this solution and suppose that <math>n = \frac{p-k}{2}</math>. Then, we have<br />
<cmath>\left(\dfrac{p-k}{2}\right)^2 \equiv -1 \mod{p}</cmath><br />
<cmath>k^2 \equiv -4 \mod{p}.</cmath><br />
Since <math>k</math> is a positive integer, it follows that <math>k^2 \ge p-4</math>, so we have<br />
<cmath>n \le \dfrac{p - \sqrt{p-4}}{2},</cmath><br />
as desired. The lemma is proven.<br />
<br />
<br />
<br />
Suppose for sake of contradiction that there are only finitely many integers <math>n_1,n_2,\dots,n_k</math> that work. Let <math>p \equiv 1 \mod{4}</math> be a prime with <math>p>20</math> and such that <math>p</math> is relatively prime to <math>n_i^2+1</math> for all <math>i</math> (the existence of <math>p</math> is guaranteed by the existence of infinitely many primes of the form <math>4k+1</math>). Then, we know that there exists an <math>N</math> such that <math>N\le \dfrac{p - \sqrt{p-4}}{2}</math> and <math>p | N^2 + 1</math> (this last condition shows that <math>N</math> is not among <math>n_1,n_2,\dots,n_k</math>. We want to show that<br />
<cmath>p > 2N + \sqrt{2N}</cmath><br />
<cmath>p - 2N > \sqrt{2N}.</cmath><br />
But, we know that <math>p-2N \ge \sqrt{p-4}</math>, so it suffices to show that<br />
<cmath>p-4 > 2N</cmath><br />
<cmath>p - 2N > 4.</cmath><br />
Once again, it suffices to show that<br />
<cmath>\sqrt{p-4} > 4,</cmath><br />
which follows from <math>p>20</math>.<br />
<br />
Thus, <math>N</math> satisfies the required condition and it follows that there exist infinitely many values of <math>n</math> that satisfy the given condition, as desired.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2008_IMO_Problems&diff=1473472008 IMO Problems2021-02-18T04:30:09Z<p>Negia: </p>
<hr />
<div>Problems of the 49th [[IMO]] 2008 Spain.<br />
<br />
== Day 1 ==<br />
<br />
=== Problem 1 ===<br />
Let <math>H</math> be the orthocenter of an acute-angled triangle <math>ABC</math>. The circle <math>\Gamma_{A}</math> centered at the midpoint of <math>BC</math> and passing through <math>H</math> intersects line <math>BC</math> at points <math>A_{1}</math> and <math>A_{2}</math>. Similarly, define the points <math>B_{1}</math>, <math>B_{2}</math>, <math>C_{1}</math> and <math>C_{2}</math>.<br />
<br />
Prove that six points <math>A_{1}</math> , <math>A_{2}</math>, <math>B_{1}</math>, <math>B_{2}</math>, <math>C_{1}</math> and <math>C_{2}</math> are concyclic.<br />
<br />
[[2008 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2 ===<br />
Let <math>x, y, z\neq 1</math> be three real numbers, such that <math>xyz = 1</math><br />
<br />
'''(i)''' Prove that;<br />
<math>\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1</math>.<br />
<br />
'''(ii)''' Prove that <math>\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} = 1</math> for infinitely many triples of rational numbers <math>x</math>, <math>y</math> and <math>z</math>.<br />
<br />
[[2008 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3 ===<br />
Prove that there are infinitely many positive integers <math>n</math> such that <math>n^{2} + 1</math> has a prime divisor greater than <math>2n + \sqrt {2n}</math>.<br />
<br />
[[2008 IMO Problems/Problem 3 | Solution]]<br />
<br />
== Day 2 ==<br />
<br />
=== Problem 4 ===<br />
Find all functions <math>f: (0, \infty) \mapsto (0, \infty)</math> (so <math>f</math> is a function from the positive real numbers) such that<br />
<center><br />
<math>\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}</math><br />
</center><br />
for all positive real numbes <math>w,x,y,z,</math> satisfying <math>wx = yz.</math><br />
<br />
[[2008 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5 ===<br />
Let <math>n</math> and <math>k</math> be positive integers with <math>k \geq n</math> and <math>k - n</math> an even number. Let <math>2n</math> lamps labelled <math>1, 2, \dots, 2n</math> be given, each of which can be either ''on'' or ''off''. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched (from on to off or from off to on).<br />
<br />
Let <math>N</math> be the number of such sequences consisting of <math>k</math> steps and resulting in the state where lamps <math>1</math> through <math>n</math> are all on, and lamps <math>n + 1</math> through <math>2n</math> are all off.<br />
<br />
Let <math>M</math> be number of such sequences consisting of <math>k</math> steps, resulting in the state where lamps <math>1</math> through <math>n</math> are all on, and lamps <math>n + 1</math> through <math>2n</math> are all off, but where none of the lamps <math>n + 1</math> through <math>2n</math> is ever switched on.<br />
<br />
Determine <math>\frac {N}{M}</math>.<br />
<br />
[[2008 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6 ===<br />
Let <math>ABCD</math> be a convex quadrilateral with <math>BA</math> different from <math>BC</math>. Denote the incircles of triangles <math>ABC</math> and <math>ADC</math> by <math>k_{1}</math> and <math>k_{2}</math> respectively. Suppose that there exists a circle <math>k</math> tangent to ray <math>BA</math> beyond <math>A</math> and to the ray <math>BC</math> beyond <math>C</math>, which is also tangent to the lines <math>AD</math> and <math>CD</math>. <br />
<br />
Prove that the common external tangents to <math>k_{1}</math> and <math>k_{2}</math> intersect on <math>k</math>.<br />
<br />
[[2008 IMO Problems/Problem 6 | Solution]]<br />
<br />
== Resources ==<br />
<br />
* [[2008 IMO]]<br />
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=2008 IMO 2008 Problems on the Resources page]<br />
<br />
{{IMO box|year=2008|before=[[2007 IMO Problems]]|after=[[2009 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2008_IMO_Problems&diff=1473462008 IMO Problems2021-02-18T04:24:08Z<p>Negia: /* Problem 2 */</p>
<hr />
<div>Problems of the 49th [[IMO]] 2008 Spain.<br />
<br />
== Day I ==<br />
<br />
=== Problem 1 ===<br />
Let <math>H</math> be the orthocenter of an acute-angled triangle <math>ABC</math>. The circle <math>\Gamma_{A}</math> centered at the midpoint of <math>BC</math> and passing through <math>H</math> intersects line <math>BC</math> at points <math>A_{1}</math> and <math>A_{2}</math>. Similarly, define the points <math>B_{1}</math>, <math>B_{2}</math>, <math>C_{1}</math> and <math>C_{2}</math>.<br />
<br />
Prove that six points <math>A_{1}</math> , <math>A_{2}</math>, <math>B_{1}</math>, <math>B_{2}</math>, <math>C_{1}</math> and <math>C_{2}</math> are concyclic.<br />
<br />
[[2008 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2 ===<br />
Let <math>x, y, z\neq 1</math> be three real numbers, such that <math>xyz = 1</math><br />
<br />
'''(i)''' Prove that;<br />
<math>\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} \geq 1</math>.<br />
<br />
'''(ii)''' Prove that <math>\frac {x^{2}}{\left(x - 1\right)^{2}} + \frac {y^{2}}{\left(y - 1\right)^{2}} + \frac {z^{2}}{\left(z - 1\right)^{2}} = 1</math> for infinitely many triples of rational numbers <math>x</math>, <math>y</math> and <math>z</math>.<br />
<br />
[[2008 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3 ===<br />
Prove that there are infinitely many positive integers <math>n</math> such that <math>n^{2} + 1</math> has a prime divisor greater than <math>2n + \sqrt {2n}</math>.<br />
<br />
[[2008 IMO Problems/Problem 3 | Solution]]<br />
<br />
== Day II ==<br />
<br />
=== Problem 4 ===<br />
Find all functions <math>f: (0, \infty) \mapsto (0, \infty)</math> (so <math>f</math> is a function from the positive real numbers) such that<br />
<center><br />
<math>\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}</math><br />
</center><br />
for all positive real numbes <math>w,x,y,z,</math> satisfying <math>wx = yz.</math><br />
<br />
[[2008 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5 ===<br />
Let <math>n</math> and <math>k</math> be positive integers with <math>k \geq n</math> and <math>k - n</math> an even number. Let <math>2n</math> lamps labelled <math>1</math>, <math>2</math>, ..., <math>2n</math> be given, each of which can be either ''on'' or ''off''. Initially all the lamps are off. We consider sequences of steps: at each step one of the lamps is switched (from on to off or from off to on).<br />
<br />
Let <math>N</math> be the number of such sequences consisting of <math>k</math> steps and resulting in the state where lamps <math>1</math> through <math>n</math> are all on, and lamps <math>n + 1</math> through <math>2n</math> are all off.<br />
<br />
Let <math>M</math> be number of such sequences consisting of <math>k</math> steps, resulting in the state where lamps <math>1</math> through <math>n</math> are all on, and lamps <math>n + 1</math> through <math>2n</math> are all off, but where none of the lamps <math>n + 1</math> through <math>2n</math> is ever switched on.<br />
<br />
Determine <math>\frac {N}{M}</math>.<br />
<br />
[[2008 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6 ===<br />
Let <math>ABCD</math> be a convex quadrilateral with <math>BA</math> different from <math>BC</math>. Denote the incircles of triangles <math>ABC</math> and <math>ADC</math> by <math>k_{1}</math> and <math>k_{2}</math> respectively. Suppose that there exists a circle <math>k</math> tangent to ray <math>BA</math> beyond <math>A</math> and to the ray <math>BC</math> beyond <math>C</math>, which is also tangent to the lines <math>AD</math> and <math>CD</math>. <br />
<br />
Prove that the common external tangents to <math>k_{1}</math> and <math>k_{2}</math> intersect on <math>k</math>.<br />
<br />
[[2008 IMO Problems/Problem 6 | Solution]]<br />
<br />
== Resources ==<br />
<br />
* [[2008 IMO]]<br />
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=2008 IMO 2008 Problems on the Resources page]<br />
<br />
{{IMO box|year=2008|before=[[2007 IMO Problems]]|after=[[2009 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2011_IMO_Problems&diff=1473452011 IMO Problems2021-02-18T04:12:35Z<p>Negia: /* Day 1 */</p>
<hr />
<div>Problems of the 52st [[IMO]] 2011 in Amsterdam, Netherlands. <br />
<br />
== Day 1 ==<br />
=== Problem 1 ===<br />
Given any set <math>A = \{a_1, a_2, a_3, a_4\}</math> of four distinct positive integers, let <math>s(A)=a_1+a_2+a_3+a_4</math> be the sum of the positive integers in <math>A</math>. Let <math>n(A)</math> denote the number of pairs <math>(i,j)</math> with <math>1 \le i < j \le 4</math> for which <math>a_i+a_j</math> divides <math>s(A)</math>. Find all sets <math>A</math> of four distinct positive integers which achieve the largest possible value of <math>n(A)</math>.<br />
<br />
''Author: Fernando Campos, Mexico''<br />
<br />
[[2011 IMO Problems/Problem 1 | Solution]]<br />
<br />
=== Problem 2 === <br />
Let <math>S</math> be a finite set of at least two points in the plane. Assume that no three points of <math>S</math> are collinear. A windmill is a process that starts with a line <math>l</math> going through a single point <math>P \in S</math>. The line rotates clockwise about the pivot <math>P</math> until the first time that the line meets some other point belonging to <math>S</math>. This point, <math>Q</math>, takes over as the new pivot, and the line now rotates clockwise about <math>Q</math>, until it next meets a point of <math>S</math>. This process continues indefinitely.<br />
Show that we can choose a point <math>P</math> in <math>S</math> and a line <math>l</math> going through <math>P</math> such that the resulting windmill uses each point of <math>S</math> as a pivot infinitely many times.<br />
<br />
''Author: Geoffrey Smith, United Kingdom''<br />
<br />
[[2011 IMO Problems/Problem 2 | Solution]]<br />
<br />
=== Problem 3 ===<br />
Let <math>f : \mathbb{R} \rightarrow \mathbb{R}</math> be a real-valued function defined on the set of real numbers that satisfies<br />
<math>f(x + y) \le yf(x) + f(f(x))</math> for all real numbers <math>x</math> and <math>y</math>. Prove that <math>f(x)=0</math> for all <math>x \le 0</math>.<br />
<br />
''Author: Igor Voronovich, Belarus''<br />
<br />
[[2011 IMO Problems/Problem 3 | Solution]]<br />
<br />
== Day 2 ==<br />
=== Problem 4. ===<br />
Let <math>n > 0</math> be an integer. We are given a balance and <math>n</math> weights of weight <math>2^0, 2^1,\ldots, 2^{n-1}</math> . We are to place each of the <math>n</math> weights on the balance, one after another, in such a way that the right pan is never heavier than the left pan. At each step we choose one of the weights that has not yet been placed on the balance, and place it on either the left pan or the right pan, until all of the weights have been placed.<br />
Determine the number of ways in which this can be done.<br />
<br />
''Author: Morteza Saghafian, Iran''<br />
<br />
[[2011 IMO Problems/Problem 4 | Solution]]<br />
<br />
=== Problem 5. ===<br />
Let <math>f</math> be a function from the set of integers to the set of positive integers. Suppose that, for any two integers <math>m</math> and <math>n</math>, the difference <math>f(m) - f(n)</math> is divisible by <math>f(m - n)</math>. Prove that, for all integers <math>m</math> and <math>n</math> with <math>f(m) \le f(n)</math>, the number <math>f(n)</math> is divisible by <math>f(m)</math>.<br />
<br />
''Author: Mahyar Sefidgaran, Iran''<br />
<br />
[[2011 IMO Problems/Problem 5 | Solution]]<br />
<br />
=== Problem 6. ===<br />
Let <math>ABC</math> be an acute triangle with circumcircle <math>\Gamma</math>. Let <math>l</math> be a tangent line to <math>\Gamma</math>, and let <math>l_a</math>, <math>l_b</math> and <math>l_c</math> be the lines obtained by reflecting <math>l</math> in the lines <math>BC</math>, <math>CA</math> and <math>AB</math>, respectively. Show that the circumcircle of the triangle determined by the lines <math>l_a</math>, <math>l_b</math> and <math>l_c</math> is tangent to the circle <math>\Gamma</math>.<br />
<br />
''Author: Japan''<br />
<br />
[[2011 IMO Problems/Problem 6 | Solution]]<br />
<br />
<br />
== Resources ==<br />
* [[2011 IMO]]<br />
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=2011&sid=8fa5e6d20f9aad1a4dd4efa73519b417 2011 IMO Problems on the Resources page]<br />
<br />
{{IMO box|year=2011|before=[[2010 IMO Problems]]|after=[[2012 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_IMO_Problems&diff=1472782013 IMO Problems2021-02-17T09:43:28Z<p>Negia: /* Problem 1 */</p>
<hr />
<div>==Problem 1==<br />
Prove that for any pair of positive integers <math>k</math> and <math>n</math>, there exist <math>k</math> positive integers <math>m_1,m_2,...,m_k</math> (not necessarily different) such that<br />
<br />
<math>1+\frac{2^k-1}{n}=\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)\cdots\left(1+\frac{1}{m_k}\right).</math><br />
<br />
[[2013 IMO Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
A configuration of <math>4027</math> points in the plane is called ''Colombian'' if it consists of <math>2013</math> red points and <math>2014</math> blue points, and no three of the points of the configuration are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is ''good'' for a Colombian<br />
configuration if the following two conditions are satisfied:<br />
*no line passes through any point of the configuration;<br />
*no region contains points of both colours.<br />
Find the least value of <math>k</math> such that for any Colombian configuration of <math>4027</math> points, there is a good<br />
arrangement of <math>k</math> lines.<br />
<br />
[[2013 IMO Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
Let the excircle of triangle <math>ABC</math> opposite the vertex <math>A</math> be tangent to the side <math>BC</math> at the point <math>A_1</math>. Define the points <math>B_1</math> on <math>CA</math> and <math>C_1</math> on <math>AB</math> analogously, using the excircles opposite <math>B</math> and <math>C</math>, respectively. Suppose that the circumcentre of triangle <math>A_1B_1C_1</math> lies on the circumcircle of triangle <math>ABC</math>. Prove that triangle <math>ABC</math> is right-angled.<br />
<br />
[[2013 IMO Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
Let <math>ABC</math> be an acute triangle with orthocenter <math>H</math>, and let <math>W</math> be a point on the side <math>BC</math>, lying strictly between <math>B</math> and <math>C</math>. The points <math>M</math> and <math>N</math> are the feet of the altitudes from <math>B</math> and <math>C</math>, respectively. Denote by <math>\omega_1</math> is [sic] the circumcircle of <math>BWN</math>, and let <math>X</math> be the point on <math>\omega_1</math> such that <math>WX</math> is a diameter of <math>\omega_1</math>. Analogously, denote by <math>\omega_2</math> the circumcircle of triangle <math>CWM</math>, and let <math>Y</math> be the point such that <math>WY</math> is a diameter of <math>\omega_2</math>. Prove that <math>X, Y</math> and <math>H</math> are collinear.<br />
<br />
[[2013 IMO Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
Let <math>\mathbb Q_{>0}</math> be the set of all positive rational numbers. Let <math>f:\mathbb Q_{>0}\to\mathbb R</math> be a function satisfying the following three conditions:<br />
<br />
(i) for all <math>x,y\in\mathbb Q_{>0}</math>, we have <math>f(x)f(y)\geq f(xy)</math>;<br />
(ii) for all <math>x,y\in\mathbb Q_{>0}</math>, we have <math>f(x+y)\geq f(x)+f(y)</math>;<br />
(iii) there exists a rational number <math>a>1</math> such that <math>f(a)=a</math>.<br />
<br />
Prove that <math>f(x)=x</math> for all <math>x\in\mathbb Q_{>0}</math>.<br />
<br />
[[2013 IMO Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
Let <math>n \ge 3</math> be an integer, and consider a circle with <math>n + 1</math> equally spaced points marked on it. Consider all labellings of these points with the numbers <math>0, 1, ... , n</math> such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels <math>a < b < c < d</math> with <math>a + d = b + c</math>, the chord joining the points labelled <math>a</math> and <math>d</math> does not intersect the chord joining the points labelled <math>b</math> and <math>c</math>.<br />
<br />
Let <math>M</math> be the number of beautiful labelings, and let N be the number of ordered pairs <math>(x, y)</math> of positive integers such that <math>x + y \le n</math> and <math>\gcd(x, y) = 1</math>. Prove that <cmath>M = N + 1.</cmath><br />
<br />
[[2013 IMO Problems/Problem 6|Solution]]<br />
<br />
{{IMO box|year=2013|before=[[2012 IMO Problems]]|after=[[2014 IMO Problems]]}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2015_IMO_Problems&diff=1470582015 IMO Problems2021-02-16T02:33:00Z<p>Negia: /* Problem 5 */</p>
<hr />
<div>==Problem 1==<br />
We say that a finite set <math>\mathcal{S}</math> in the plane is <i> balanced </i><br />
if, for any two different points <math>A</math>, <math>B</math> in <math>\mathcal{S}</math>, there is<br />
a point <math>C</math> in <math>\mathcal{S}</math> such that <math>AC=BC</math>. We say that<br />
<math>\mathcal{S}</math> is <i>centre-free</i> if for any three points <math>A</math>, <math>B</math>, <math>C</math> in<br />
<math>\mathcal{S}</math>, there is no point <math>P</math> in <math>\mathcal{S}</math> such that<br />
<math>PA=PB=PC</math>.<br />
<ol style="list-style-type: lower-latin;"><br />
<li> Show that for all integers <math>n\geq 3</math>, there exists a balanced set consisting of <math>n</math> points. </li><br />
<li> Determine all integers <math>n\geq 3</math> for which there exists a balanced centre-free set consisting of <math>n</math> points. </li> <br />
</ol><br />
<br />
[[2015 IMO Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
Determine all triples of positive integers <math>(a,b,c)</math> such that each of the numbers <br />
<cmath> ab-c,\; bc-a,\; ca-b </cmath><br />
is a power of 2. <br />
<br />
(<i>A power of 2 is an integer of the form <math>2^n</math> where <math>n</math> is a non-negative integer </i>).<br />
<br />
[[2015 IMO Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
Let <math>ABC</math> be an acute triangle with <math>AB>AC</math>. Let <math>\Gamma</math> be its circumcircle, <math>H</math> its orthocenter, and <math>F</math> the foot of the altitude from <math>A</math>. Let <math>M</math> be the midpoint of <math>BC</math>. Let <math>Q</math> be the point on <math>\Gamma</math> such that <math>\angle HKQ=90^\circ</math>. Assume that the points <math>A</math>, <math>B</math>, <math>C</math>, <math>K</math>, and <math>Q</math> are all different, and lie on <math>\Gamma</math> in this order.<br />
<br />
Prove that the circumcircles of triangles <math>KQH</math> and <math>FKM</math> are tangent to each other.<br />
<br />
[[2015 IMO Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
Triangle <math>ABC</math> has circumcircle <math>\Omega</math> and circumcenter <math>O</math>. A circle <math>\Gamma</math> with center <math>A</math> intersects the segment <math>BC</math> at points <math>D</math> and <math>E</math>, such that <math>B</math>, <math>D</math>, <math>E</math>, and <math>C</math> are all different and lie on line <math>BC</math> in this order. Let <math>F</math> and <math>G</math> be the points of intersection of <math>\Gamma</math> and <math>\Omega</math>, such that <math>A</math>, <math>F</math>, <math>B</math>, <math>C</math>, and <math>G</math> lie on <math>\Omega</math> in this order. Let <math>K</math> be the second point of intersection of the circumcircle of triangle <math>BDF</math> and the segment <math>AB</math>. Let <math>L</math> be the second point of intersection of the circumcircle of triangle <math>CGE</math> and the segment <math>CA</math>.<br /><br /><br />
Suppose that the lines <math>FK</math> and <math>GL</math> are different and intersect at the point <math>X</math>. Prove that <math>X</math> lies on the line <math>AO</math>.<br /><br /><br />
<br />
[[2015 IMO Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
Let <math>\mathbb{R}</math> be the set of real numbers. Determine all functions <math>f:\mathbb{R}\rightarrow\mathbb{R}</math> satisfying the equation<br />
<br />
<math>f(x+f(x+y))+f(xy) = x+f(x+y)+yf(x)</math><br />
<br />
for all real numbers <math>x</math> and <math>y</math>.<br />
<br />
[[2015 IMO Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
The sequence <math>a_1,a_2,\dots</math> of integers satisfies the conditions:<br /><br /><br />
(i) <math>1\le a_j\le2015</math> for all <math>j\ge1</math>,<br /><br />
(ii) <math>k+a_k\neq \ell+a_\ell</math> for all <math>1\le k<\ell</math>.<br /><br /><br />
Prove that there exist two positive integers <math>b</math> and <math>N</math> for which<cmath>\left\vert\sum_{j=m+1}^n(a_j-b)\right\vert\le1007^2</cmath>for all integers <math>m</math> and <math>n</math> such that <math>n>m\ge N</math>.<br /><br /><br />
<br />
[[2015 IMO Problems/Problem 6|Solution]]</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2001_IMO_Problems&diff=1470362001 IMO Problems2021-02-16T02:26:28Z<p>Negia: /* Problem 6 */</p>
<hr />
<div>==Problem 1==<br />
<br />
Consider an acute triangle <math>\triangle ABC</math>. Let <math>P</math> be the foot of the altitude of triangle <math>\triangle ABC</math> issuing from the vertex <math>A</math>, and let <math>O</math> be the [[circumcenter]] of triangle <math>\triangle ABC</math>. Assume that <math>\angle C \geq \angle B+30^{\circ}</math>. Prove that <math>\angle A+\angle COP < 90^{\circ}</math>.<br />
<br />
==Problem 2==<br />
<br />
Let <math>a,b,c</math> be positive real numbers. Prove that <math>\frac{a}{\sqrt{a^{2}+8bc}}+\frac{b}{\sqrt{b^{2}+8ca}}+\frac{c}{\sqrt{c^{2}+8ab}}\ge 1</math>.<br />
<br />
==Problem 3==<br />
<br />
Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that each contestant solved at most six problems, and for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy. Show that there is a problem that was solved by at least three girls and at least three boys.<br />
<br />
==Problem 4==<br />
<br />
Let <math>n_1, n_2, \dots , n_m</math> be integers where <math>m>1</math> is odd. Let <math>x = (x_1, \dots , x_m)</math> denote a permutation of the integers <math>1, 2, \cdots , m</math>. Let <math>f(x) = x_1n_1 + x_2n_2 + ... + x_mn_m</math>. Show that for some distinct permutations <math>a</math>, <math>b</math> the difference <math>f(a) - f(b)</math> is a multiple of <math>m!</math>.<br />
<br />
==Problem 5==<br />
<br />
<math>ABC</math> is a [[triangle]]. <math>X</math> lies on <math>BC</math> and <math>AX</math> bisects [[angle]] <math>A</math>. <math>Y</math> lies on <math>CA</math> and <math>BY</math> bisects angle <math>B</math>. Angle <math>A</math> is <math>60^{\circ}</math>. <math>AB + BX = AY + YB</math>. Find all possible values for angle <math>B</math>.<br />
<br />
==Problem 6==<br />
<br />
<math>k > l > m > n</math> are positive integers such that <math>km + ln = (k+l-m-n)(-k+l+m+n)</math>. Prove that <math>kl+mn</math> is not prime.<br />
<br />
==See Also==<br />
<br />
* [[2001 IMO]]<br />
* [[IMO Problems and Solutions]]<br />
* [[IMO]]</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1409242013 USAMO Problems/Problem 12020-12-29T14:30:36Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We will use construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
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.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,<br />
<cmath>\begin{align*}<br />
\angle BRY'+\angle BPY'<br />
&=180^\circ-\theta+\theta\\<br />
&=180^\circ.<br />
\end{align*}</cmath><br />
<br />
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.<br />
<br />
Since points <math>Y'</math> and <math>Z'</math> lie on <math>AP,</math> we know that,<br />
<cmath>Y'=\omega_B\cap AP</cmath><br />
and that<br />
<cmath>Z'=\omega_C\cap AP.</cmath><br />
<br />
Hence, the points <math>Y'</math> and <math>Z'</math> coincide with the given points <math>Y</math> and <math>Z,</math> respectively.<br />
<br />
Since quadrilateral <math>AQTR</math> is also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle Y'TZ'<br />
&=180^\circ-\angle RTQ\\<br />
&=180^\circ-(180^\circ-\angle RAQ)\\<br />
&=\angle RAQ\\<br />
&=\alpha.<br />
\end{align*}</cmath><br />
<br />
Similarly, since quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle TY'Z'<br />
&=180^\circ-\angle RY'P\\<br />
&=180^\circ-(180^\circ-\angle RBP)\\<br />
&=\angle RBP\\<br />
&=\beta,<br />
\end{align*}</cmath><br />
and,<br />
<cmath>\begin{align*}<br />
\angle Y'Z'T<br />
&=180^\circ-\angle PZ'Q\\<br />
&=180^\circ-(180^\circ-\angle PCQ)\\<br />
&=\angle PCQ\\<br />
&=\gamma.<br />
\end{align*}</cmath><br />
<br />
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><br />
<br />
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.<br />
<br />
Hence, quadrilateral <math>AQTX</math> is cyclic.<br />
<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564));<br />
draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
Since the angles <math>\angle ART</math> and <math>\angle AXT</math> are inscribed in the same arc <math>\overarc{AT},</math> we have,<br />
<cmath>\begin{align*}<br />
\angle AXT<br />
&=\angle ART\\<br />
&=\theta.<br />
\end{align*}</cmath><br />
<br />
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,<br />
<cmath>Y'X/XZ'=BP/PC.</cmath><br />
<br />
Since <math>Y'=Y</math> and <math>Z'=Z</math> hence,<br />
<cmath>YX/XZ=BP/PC,</cmath><br />
<br />
as required.<br />
<br />
-Negia<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1409232013 USAMO Problems/Problem 12020-12-29T14:30:07Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We will use construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
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.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,<br />
<cmath>\begin{align*}<br />
\angle BRY'+\angle BPY'<br />
&=180^\circ-\theta+\theta\\<br />
&=180^\circ.<br />
\end{align*}</cmath><br />
<br />
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.<br />
<br />
Since points <math>Y'</math> and <math>Z'</math> lie on <math>AP,</math> we know that,<br />
<cmath>Y'=\omega_B\cap AP</cmath><br />
and that<br />
<cmath>Z'=\omega_C\cap AP.</cmath><br />
<br />
Hence, the points <math>Y'</math> and <math>Z'</math> coincide with the given points <math>Y</math> and <math>Z,</math> respectively.<br />
<br />
Since quadrilateral <math>AQTR</math> is also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle Y'TZ'<br />
&=180^\circ-\angle RTQ\\<br />
&=180^\circ-(180^\circ-\angle RAQ)\\<br />
&=\angle RAQ\\<br />
&=\alpha.<br />
\end{align*}</cmath><br />
<br />
Similarly, since quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle TY'Z'<br />
&=180^\circ-\angle RY'P\\<br />
&=180^\circ-(180^\circ-\angle RBP)\\<br />
&=\angle RBP\\<br />
&=\beta,<br />
\end{align*}</cmath><br />
and,<br />
<cmath>\begin{align*}<br />
\angle Y'Z'T<br />
&=180^\circ-\angle PZ'Q\\<br />
&=180^\circ-(180^\circ-\angle PCQ)\\<br />
&=\angle PCQ\\<br />
&=\gamma.<br />
\end{align*}</cmath><br />
<br />
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><br />
<br />
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.<br />
<br />
Hence, quadrilateral <math>AQTX</math> is cyclic.<br />
<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564));<br />
draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
Since the angles <math>\angle ART</math> and <math>\angle AXT</math> are inscribed in the same arc <math>\overarc{AT},</math> we have,<br />
<cmath>\begin{align*}<br />
\angle AXT<br />
&=\angle ART\\<br />
&=\theta.<br />
\end{align*}</cmath><br />
<br />
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,<br />
<cmath>Y'X/XZ'=BP/PC.</cmath><br />
<br />
Since <math>Y'=Y</math> and <math>Z'=Z</math> hence,<br />
<cmath>YX/XZ=BP/PC,</cmath><br />
<br />
as required.<br />
-negia<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=1994_USAMO_Problems/Problem_1&diff=1392261994 USAMO Problems/Problem 12020-12-08T10:09:15Z<p>Negia: Added a 3rd Solution</p>
<hr />
<div>==Problem==<br />
Let <math> \, k_1 < k_2 < k_3 <\cdots\, </math>, be positive integers, no two consecutive, and let <math> \, s_m = k_1+k_2+\cdots+k_m\, </math>, for <math> \, m = 1,2,3,\ldots\;\; </math>. Prove that, for each positive integer <math>n</math>, the interval <math> \, [s_n, s_{n+1})\, </math>, contains at least one perfect square.<br />
<br />
==Solution==<br />
We want to show that the distance between <math>s_n</math> and <math>s_{n+1}</math> is greater than the distance between <math>s_n</math> and the next perfect square following <math>s_n</math>.<br />
<br />
Given <math>s_n=\sum_{i=1}^{n}k_i</math>, where no <math>k_i</math> are consecutive, we can put a lower bound on <math>k_n</math>. This occurs when all <math>k_{i+1}=k_i+2</math>:<br />
<br />
<cmath><br />
\begin{align*}<br />
s_n&=(k_{n,min})+(k_{n,min}-2)+(k_{n,min}-4)+\dots+(k_{n,min}-2n+2)\\<br />
&=nk_{n,min}-\sum_{i=1}^{n-1}2i\\<br />
&=nk_{n,min}-2\sum_{i=1}^{n}i+2n\\<br />
&=nk_{n,min}-n(n+1)+2n\\<br />
&=nk_{n,min}-n^2+n\\<br />
\end{align*}<br />
</cmath><br />
<br />
Rearranging, <math>k_{n,min}=\frac{s_n}{n}+n-1</math>. So, <math>k_n\geq\frac{s_n}{n}+n-1</math>, and the distance between <math>s_n</math> and <math>s_{n+1}</math> is <math>k_{n+1}\geq k_n+2\geq\frac{s_n}{n}+n+1</math>.<br />
<br />
Also, let <math>d(s_n)</math> be the distance between <math>s_n</math> and the next perfect square following <math>s_n</math>. Let's look at the function <math>d(x)</math> for all positive integers <math>x</math>.<br />
<br />
When <math>x</math> is a perfect square, it is easy to see that <math>d(x)=2\sqrt{x}+1</math>.<br />
Proof: Choose <math>x=m^2</math>. <math>d(m^2)=(m+1)^2-m^2=2m+1=2\sqrt{m^2}+1</math>.<br />
<br />
When <math>x</math> is not a perfect square, <math>d(x)<2\sqrt{x}+1</math>.<br />
Proof: Choose <math>x=m^2+p</math> with <math>0<p<2m+1</math>. <math>d(m^2+p)=(m+1)^2-m^2-p=2m+1-p<2m+1=2\sqrt{m^2}+1<2\sqrt{m^2+p}+1</math>.<br />
<br />
So, <math>d(x)\leq 2\sqrt{x}+1</math> for all <math>x</math> and <math>d(s_n)\leq 2\sqrt{s_n}+1</math> for all <math>s_n</math>.<br />
<br />
Now, it suffices to show that <math>k_{n+1}\geq d(s_n)</math> for all <math>n</math>.<br />
<br />
<cmath><br />
\begin{align*}<br />
k_{n+1}-d(s_n)&\geq \frac{s_n}{n}+n+1-2\sqrt{s_n}-1\\<br />
&=\frac{1}{n}(s_n+n^2-2n\sqrt{s_n})\\<br />
&=\frac{s_n^2+n^4+2n^2s_n-4n^2s_n}{n(s_n+n^2+2n\sqrt{s_n})}\\<br />
&=\frac{(s_n-n^2)^2}{n(s_n+n^2+2n\sqrt{s_n})}\\<br />
&\geq 0<br />
\end{align*}<br />
</cmath><br />
<br />
So, <math>k_{n+1}\geq d(s_n)</math> and all intervals between <math>s_n</math> and <math>s_{n+1}</math> will contain at least one perfect square.<br />
<br />
==Solution 2==<br />
<br />
We see that by increasing <math>n</math> by some amount, we simply shift our interval by a finite amount. It suffices to consider the case <math>n=1</math> (since this can be inducted across all positive integers). Let <math>k_1=x</math>. We want the smallest interval, so we have <math>[x, 2x+2]</math>. Simple induction reveals that the ration of consecutive squares grows slower than our linear bound. We now consider sufficiently small <math>x</math> (where <math>\frac{(n+1)^2}{n^2}<2</math>). This first happens at <math>n=3</math>. By simple casework, our answer is as desired <math>\blacksquare</math>.<br />
<br />
<br />
==Solution 3==<br />
We will first prove by Induction on <math>n\in\mathbb{N}</math> that <math>(k_{n}+1)^2\geq4(k_1+k_2+\cdots +k_{n}).</math> Denote this statement by <math>P(n).</math><br />
<br />
For the Base Case let <math>n=1,</math> we know that;<br />
<cmath>(k_1-1)^2\geq 0</cmath><br />
<cmath>\Rightarrow k_1^2-2k_1+1\geq 0</cmath><br />
<cmath>\Rightarrow k_1^2+2k_1+1\geq 4k_1</cmath><br />
<cmath>\Rightarrow (k_1+1)^2\geq 4k_1.</cmath><br />
<br />
Hence, the Base Case holds.<br />
<br />
For the Inductive Step, suppose that <math>P(m)</math> holds for some <math>k\in\mathbb{N},</math> we will prove that <math>P(m+1)</math> holds as well.<br />
<br />
Assume for contradiction that <math>P(m+1)</math> doesn't hold, then we know that;<br />
<cmath>(k_m+1)^2<4(k_1+k_2+\cdots +k_m)</cmath><br />
<cmath>\Rightarrow k_m^2+2k_m+1<4(k_1+k_2+\cdots +k_{m-1})+4k_m</cmath><br />
<cmath>\Rightarrow k_m^2-2k_m+1<4(k_1+k_2+\cdots +k_{m-1})</cmath><br />
<cmath>\Rightarrow (k_m-1)^2<4(k_1+k_2+\cdots +k_{m-1}).</cmath><br />
<br />
We know that since <math>k_m</math> and <math>k_{m-1}</math> are not consecutive, we have;<br />
<cmath>k_{m-1}\geq k_m-2</cmath><br />
<cmath>\Rightarrow k_{m-1}+1\leq k_m-1</cmath><br />
<cmath>\Rightarrow (k_{m-1}+1)^2\leq (k_m-1)^2</cmath><br />
<cmath>\Rightarrow (k_{m-1}+1)^2\leq (k_m-1)^2<4(k_1+k_2+\cdots +k_{m-1}).</cmath><br />
<br />
But this contradicts the Inductive Hypothesis that <cmath>\Rightarrow (k_{m-1}+1)^2\geq 4(k_1+k_2+\cdots +k_{m-1}),</cmath> thus our assumption was false and the Inductive Step is complete.<br />
<br />
Hence, we have proved that <math>P(n)</math> holds, and we will use <math>P(n)</math> to solve the original problem.<br />
<br />
<br />
Now suppose that for some positive integer <math>n</math>, the interval <math>\, [s_n, s_{n+1})\,</math> does not contain any perfect square, then we know that there must exist two perfect squares of consecutive integers, such that the smaller one is lesser than <math>s_n</math> and the larger one greater than or equal to <math>s_{n+1}.</math><br />
<br />
We thus know that there exists some <math>x\in\mathbb{N}</math> such that;<br />
<cmath><br />
\begin{eqnarray}<br />
x^2<s_n \\<br />
\text{and }(x+1)^2\geq s_{n+1}<br />
\end{eqnarray}<br />
</cmath><br />
<br />
By Inequality 1;<br />
<cmath><br />
\begin{align*}<br />
x<br />
&<\sqrt{s_n}\\<br />
&=\sqrt{k_1+k_2+\cdots+k_n}.\\<br />
\end{align*}<br />
</cmath><br />
<br />
<br />
Hence, we know that;<br />
<cmath><br />
\begin{align*}<br />
(x+1)^2<br />
&<(\sqrt{k_1+k_2+\cdots+k_n}+1)^2\\<br />
&=k_1+k_2+\cdots+k_n+2\sqrt{k_1+k_2+\cdots+k_n}+1.<br />
\end{align*}<br />
</cmath><br />
<br />
Combining this with Inequality 2 gives;<br />
<cmath>s_{n+1}\leq (x+1)^2<k_1+k_2+\cdots+k_n+2\sqrt{k_1+k_2+\cdots+k_n}+1</cmath><br />
<cmath>\Rightarrow k_1+k_2+\cdots+k_n+k_{n+1}<k_1+k_2+\cdots+k_n+2\sqrt{k_1+k_2+\cdots+k_n}+1</cmath><br />
<cmath>\Rightarrow k_{n+1}<2\sqrt{k_1+k_2+\cdots+k_n}+1</cmath><br />
<cmath>\Rightarrow k_{n+1}-1<2\sqrt{k_1+k_2+\cdots+k_n}.</cmath><br />
<br />
<br />
We know that since <math>k_{n+1}</math> are not consecutive, <math>k_n+1\leq k_{n+1}-1,</math> hence, we have;<br />
<cmath>k_{n+1}\leq k_{n+1}-1<2\sqrt{k_1+k_2+\cdots+k_n}</cmath><br />
<cmath>k_{n+1}<2\sqrt{k_1+k_2+\cdots+k_n}</cmath><br />
<cmath>(k_{n+1}^2<4(k_1+k_2+\cdots+k_n).</cmath><br />
<br />
<br />
But this contradicts the statement <math>P(n)</math> which was proved earlier.<br />
<math>\square</math><br />
==See Also==<br />
{{USAMO box|year=1994|before=First Problem|num-a=2}}<br />
{{MAA Notice}}<br />
[[Category:Olympiad Number Theory Problems]]</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=1998_USAMO_Problems&diff=1383541998 USAMO Problems2020-11-24T10:11:08Z<p>Negia: /* Problem 3 */</p>
<hr />
<div>Problems of the [[1998 USAMO | 1998]] [[USAMO]].<br />
<br />
==Day 1==<br />
===Problem 1===<br />
Suppose that the set <math>\{1,2,\cdots, 1998\}</math> has been partitioned into disjoint pairs <math>\{a_i,b_i\}</math> (<math>1\leq i\leq 999</math>) so that for all <math>i</math>, <math>|a_i-b_i|</math> equals <math>1</math> or <math>6</math>. Prove that the sum<br />
<cmath> |a_1-b_1|+|a_2-b_2|+\cdots +|a_{999}-b_{999}| </cmath><br />
ends in the digit <math>9</math>.<br />
<br />
[[1998 USAMO Problems/Problem 1|Solution]]<br />
<br />
===Problem 2===<br />
Let <math>{\cal C}_1</math> and <math>{\cal C}_2</math> be concentric circles, with <math>{\cal C}_2</math> in the interior of <math>{\cal C}_1</math>. From a point <math>A</math> on <math>{\cal C}_1</math> one draws the tangent <math>AB</math> to <math>{\cal C}_2</math> (<math>B\in {\cal C}_2</math>). Let <math>C</math> be the second point of intersection of <math>AB</math> and <math>{\cal C}_1</math>, and let <math>D</math> be the midpoint of <math>AB</math>. A line passing through <math>A</math> intersects <math>{\cal C}_2</math> at <math>E</math> and <math>F</math> in such a way that the perpendicular bisectors of <math>DE</math> and <math>CF</math> intersect at a point <math>M</math> on <math>AB</math>. Find, with proof, the ratio <math>AM/MC</math>.<br />
<br />
[[1998 USAMO Problems/Problem 2|Solution]]<br />
<br />
===Problem 3===<br />
Let <math>a_0,a_1,\cdots ,a_n</math> be numbers from the interval <math>(0,\pi/2)</math> such that<br />
<cmath> \tan \left(a_0-\frac{\pi}{4}\right)+ \tan \left(a_1-\frac{\pi}{4}\right)+\cdots +\tan \left(a_n-\frac{\pi}{4}\right)\geq n-1. </cmath><br />
Prove that<br />
<cmath> \tan a_0\tan a_1 \cdots \tan a_n\geq n^{n+1}. </cmath><br />
[[1998 USAMO Problems/Problem 3|Solution]]<br />
<br />
==Day 2==<br />
===Problem 4===<br />
A computer screen shows a <math>98 \times 98</math> chessboard, colored in the usual way. One can select with a mouse any rectangle with sides on the lines of the chessboard and click the mouse button: as a result, the colors in the selected rectangle switch (black becomes white, white becomes black). Find, with proof, the minimum number of mouse clicks needed to make the chessboard all one color.<br />
<br />
[[1998 USAMO Problems/Problem 4|Solution]]<br />
<br />
===Problem 5===<br />
Prove that for each <math>n\geq 2</math>, there is a set <math>S</math> of <math>n</math> integers such that <math>(a-b)^2</math> divides <math>ab</math> for every distinct <math>a,b\in S</math>.<br />
<br />
[[1998 USAMO Problems/Problem 5|Solution]]<br />
<br />
===Problem 6===<br />
Let <math>n \geq 5</math> be an integer. Find the largest integer <math>k</math> (as a function of <math>n</math>) such that there exists a convex <math>n</math>-gon <math>A_{1}A_{2}\dots A_{n}</math> for which exactly <math>k</math> of the quadrilaterals <math>A_{i}A_{i+1}A_{i+2}A_{i+3}</math> have an inscribed circle. (Here <math>A_{n+j} = A_{j}</math>.)<br />
<br />
[[1998 USAMO Problems/Problem 6|Solution]]<br />
<br />
== See Also ==<br />
{{USAMO newbox|year=1998|before=[[1997 USAMO]]|after=[[1999 USAMO]]}}<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=User:Rusczyk&diff=131344User:Rusczyk2020-08-10T15:37:43Z<p>Negia: /* User Count */</p>
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<div>Rusczyk's Page:<br />
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==<font color="white" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="white">Rusczyk is currently borderline AIME.<br><br />
<br />
Rusczyk just turned 13 years old.<br><br />
<br />
Rusczyk scored 44/46 when mocking the 2018 MATHCOUNTS State test, and got silver on the 2020 online MATHCOUNTS State held on AoPS.<br><br />
<br />
Rusczyk is a pro at maths and physics<br />
<br />
Rusczyk has come world and country #1 in various international tournaments and competitions starting from 2017<br />
<br />
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<br />
==<font color="white" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
<div style="margin-left: 10px; margin-right: 10px; margin-bottom:10px"><font color="white"> A User Count of 500<br />
<br />
Make AIME 2021 (Currently borderline)<br />
<br />
Pass AP Calculus AB, BC and AP Physics exam<br />
<br />
Get in the Alcumus HoF in the next 6 months<br />
<br />
Get <math>\color{white}{2 \times}</math> medals this year as compared to what they did last year. That is <math>\color{white}{2 \times 14 = \boxed{28}}</math> which is nearly impossible.<br />
</div><br />
</div></div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2015_USAMO_Problems/Problem_3&diff=1313432015 USAMO Problems/Problem 32020-08-10T15:35:22Z<p>Negia: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>S = \{1, 2, ..., n\}</math>, where <math>n \ge 1</math>. Each of the <math>2^n</math> subsets of <math>S</math> is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set <math>T \subseteq S</math>, we then write <math>f(T)</math> for the number of subsets of T that are blue. <br />
<br />
Determine the number of colorings that satisfy the following condition: for any subsets <math>T_1</math> and <math>T_2</math> of <math>S</math>, <cmath>f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2).</cmath><br />
<br />
===Solution===<br />
Define function: <math>C(T)=1</math> if the set T is colored blue, and <math>C(T)=0</math> if <math>T</math> is colored red.<br />
Define the <math>\text{Core} =\text{intersection of all } T \text{ where } C(T)=1</math>. <br />
<br />
The empty set is denoted as <math>\varnothing</math>, <math>\cap</math> denotes intersection, and <math>\cup</math> denotes union. Let <math>S_n=\{n\}</math> are one-element subsets.<br />
<br />
Let <math>m_{c_k}=\dbinom{m}{k} = \frac{m!}{k!(m-k)!}</math> denote m choose k.<br />
<br />
<br />
(Case I) <math>f(\varnothing)=1</math>. Then for distinct m and k, <math>f(S_m \cup S_k)=f(S_m)f(S_k)</math>, meaning only if <math>S_m</math> and <math>S_k</math> are both blue is their union blue. Namely <math>C(S_m \cup S_k)=C(S_m)C(S_k).</math><br />
<br />
Similarly, for distinct <math>m,n,k</math>, <math>f(S_m \cup S_k \cup Sn)=f(S_m \cup S_k)f(S_n)</math>, <math>C(S_m \cup S_k \cup S_n)=C(S_m)C(S_k)C(S_n)</math>. This procedure of determination continues to <math>S</math>. Therefore, if <math>T=\{a_1,a_2, \cdots a_k\}</math>, then <math>C(T)=C(S_{a_1})C(S_{a_2}) \cdots C(S_{a_k})</math>. All colorings thus determined by the free colors chosen for subsets of one single elements satisfy the condition. There are <math>2^n</math> colorings in this case. <br />
<br />
(Case II.) <math>f(\varnothing)=0</math>.<br />
<br />
(Case II.1) <math>\text{Core}=\varnothing</math>. Then either (II.1.1) there exist two nonintersecting subsets A and B, <math>C(A)=C(B)=1</math>, but f<math>(A)f(B)=0</math>, a contradiction, or (II.1.2) all subsets has <math>C(T)=0</math>, which is easily confirmed to satisfy the condition <math>f(T_1)f(T_2)=f(T_1 \cap T_2)f(T_1 \cup T_2)</math>. There is one coloring in this case.<br />
<br />
(Case II.2) Core = a subset of 1 element. WLOG, <math>C(S_1)=1</math>. Then <math>f(S_1)=1</math>, and subsets containing element 1 may be colored Blue. <math>f(S_1 \cup S_m)f(S_1\cup S_n)=f(S_1 \cup S_m \cup S_n)</math>, namely <math>C(S_1 \cup S_m \cup S_n)=C(S_m \cup S_1)C(S_n \cup S_1)</math>. Now S_1 functions as the <math>\varnothing</math> in case I, with <math>n-1</math> elements to combine into a base of <math>n-1</math> two-element sets, and all the other subsets are determined. There are <math>2^{n-1}</math> colorings for each choice of core. However, there are nC1 = n such cores. Hence altogether there are <math>n2^{n-1}</math> colorings in this case.<br />
<br />
(Case II.3) Core = a subset of 2 elements. WLOG, let <math>C(S_1 \cup S_2)=1</math>. Only subsets containing the core may be colored blue. With the same reasoning as in the preceding case, there are <math>(nC2)2^{n-2}</math> colorings.<br />
<br />
<math>\dots</math><br />
<br />
(Case II.n+1) Core = S. Then <math>C(S)=1</math>, with all other subsets <math>C(T)=0</math>, there is <math>1=\dbinom{n}{n}2^0</math><br />
<br />
Combining all the cases, we have <math>1+\left[2^n+\dbinom{n}{1}2^{n-1}+\dbinom{n}{2}2^{n-2}+ \cdots + \dbinom{n}{n}2^0\right]=\boxed{1+3^n}</math> colorings.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2015_USAMO_Problems/Problem_3&diff=1313392015 USAMO Problems/Problem 32020-08-10T15:09:01Z<p>Negia: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>S = \{1, 2, ..., n\}</math>, where <math>n \ge 1</math>. Each of the <math>2^n</math> subsets of <math>S</math> is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set <math>T \subseteq S</math>, we then write <math>f(T)</math> for the number of subsets of T that are blue. <br />
<br />
Determine the number of colorings that satisfy the following condition: for any subsets <math>T_1</math> and <math>T_2</math> of <math>S</math>, <cmath>f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2).</cmath><br />
<br />
===Solution===<br />
Define function: <math>C(T)=1</math> if the set T is colored blue, and <math>C(T)=0</math> if <math>T</math> is colored red.<br />
Define the <math>\text{Core} =\text{intersection of all } T \text{ where } C(T)=1</math>. <br />
<br />
The empty set is denoted as <math>\varnothing</math>, <math>\cap</math> denotes intersection, and <math>\cup</math> denotes union. Let <math>S_n=\{n\}</math> are one-element subsets.<br />
<br />
Let <math>m_{c_k}=\dbinom{m}{k} = \frac{m!}{k!(m-k)!}</math> denote m choose k.<br />
<br />
<br />
(Case I) <math>f(\varnothing)=1</math>. Then for distinct m and k, <math>f(S_m \cup S_k)=f(S_m)f(S_k)</math>, meaning only if <math>S_m</math> and <math>S_k</math> are both blue is their union blue. Namely <math>C(S_m \cup S_k)=C(S_m)C(S_k).</math><br />
<br />
Similarly, for distinct <math>m,n,k</math>, <math>f(S_m \cup S_k \cup Sn)=f(S_m \cup S_k)f(S_n)</math>, <math>C(S_m \cup S_k \cup S_n)=C(S_m)C(S_k)C(S_n)</math>. This procedure of determination continues to <math>S</math>. Therefore, if <math>T=\{a_1,a_2, \cdots a_k\}</math>, then <math>C(T)=C(S_{a_1})C(S_{a_2}) \cdots C(S_{a_k})</math>. All colorings thus determined by the free colors chosen for subsets of one single elements satisfy the condition. There are <math>2^n</math> colorings in this case. <br />
<br />
(Case II.) <math>f(\varnothing)=0</math>.<br />
<br />
(Case II.1) <math>\text{Core}=\varnothing</math>. Then either (II.1.1) there exist two nonintersecting subsets A and B, <math>C(A)=C(B)=1</math>, but f<math>(A)f(B)=0</math>, a contradiction, or (II.1.2) all subsets has <math>C(T)=0</math>, which is easily confirmed to satisfy the condition <math>f(T_1)f(T_2)=f(T_1 \cap T_2)f(T_1 \cup T_2)</math>. There is one coloring in this case.<br />
<br />
(Case II.2) Core = a subset of 1 element. WLOG, C(S_1)=1. Then <math>f(S_1)=1</math>, and subsets containing element 1 may be colored Blue. <math>f(S_1 \cup S_m)f(S_1\cup S_n)=f(S_1 \cup S_m \cup S_n)</math>, namely <math>C(S_1 \cup S_m \cup S_n)=C(S_m \cup S_1)C(S_n \cup S_1)</math>. Now S_1 functions as the <math>\varnothing</math> in case I, with <math>n-1</math> elements to combine into a base of <math>n-1</math> two-element sets, and all the other subsets are determined. There are <math>2^{n-1}</math> colorings for each choice of core. However, there are nC1 = n such cores. Hence altogether there are <math>n2^{n-1}</math> colorings in this case.<br />
<br />
(Case II.3) Core = a subset of 2 elements. WLOG, let <math>C(S_1 \cup S_2)=1</math>. Only subsets containing the core may be colored blue. With the same reasoning as in the preceding case, there are <math>(nC2)2^{n-2}</math> colorings.<br />
<br />
<math>\dots</math><br />
<br />
(Case II.n+1) Core = S. Then <math>C(S)=1</math>, with all other subsets <math>C(T)=0</math>, there is <math>1=\dbinom{n}{n}2^0</math><br />
<br />
Combining all the cases, we have <math>1+\left[2^n+\dbinom{n}{1}2^{n-1}+\dbinom{n}{2}2^{n-2}+ \cdots + \dbinom{n}{n}2^0\right]=\boxed{1+3^n}</math> colorings.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2015_USAMO_Problems/Problem_3&diff=1313382015 USAMO Problems/Problem 32020-08-10T15:08:12Z<p>Negia: /* Solution */</p>
<hr />
<div>==Problem==<br />
Let <math>S = \{1, 2, ..., n\}</math>, where <math>n \ge 1</math>. Each of the <math>2^n</math> subsets of <math>S</math> is to be colored red or blue. (The subset itself is assigned a color and not its individual elements.) For any set <math>T \subseteq S</math>, we then write <math>f(T)</math> for the number of subsets of T that are blue. <br />
<br />
Determine the number of colorings that satisfy the following condition: for any subsets <math>T_1</math> and <math>T_2</math> of <math>S</math>, <cmath>f(T_1)f(T_2) = f(T_1 \cup T_2)f(T_1 \cap T_2).</cmath><br />
<br />
===Solution===<br />
Define function: <math>C(T)=1</math> if the set T is colored blue, and <math>C(T)=0</math> if <math>T</math> is colored red.<br />
Define the <math>\text{Core} =\text{intersection of all } T \text{ where } C(T)=1</math>. <br />
<br />
The empty set is denoted as <math>\varnothing</math>, <math>\cap</math> denotes intersection, and <math>\cup</math> denotes union. Let <math>S_n=\{n\}</math> are one-element subsets.<br />
<br />
Let <math>m_{c_k}=\dbinom{m}{k} = \frac{m!}{k!(m-k)!}</math> denote m choose k.<br />
<br />
<br />
(Case I) <math>f(\null)=1</math>. Then for distinct m and k, <math>f(S_m \cup S_k)=f(S_m)f(S_k)</math>, meaning only if <math>S_m</math> and <math>S_k</math> are both blue is their union blue. Namely <math>C(S_m \cup S_k)=C(S_m)C(S_k).</math><br />
<br />
Similarly, for distinct <math>m,n,k</math>, <math>f(S_m \cup S_k \cup Sn)=f(S_m \cup S_k)f(S_n)</math>, <math>C(S_m \cup S_k \cup S_n)=C(S_m)C(S_k)C(S_n)</math>. This procedure of determination continues to <math>S</math>. Therefore, if <math>T=\{a_1,a_2, \cdots a_k\}</math>, then <math>C(T)=C(S_{a_1})C(S_{a_2}) \cdots C(S_{a_k})</math>. All colorings thus determined by the free colors chosen for subsets of one single elements satisfy the condition. There are <math>2^n</math> colorings in this case. <br />
<br />
(Case II.) <math>f(\varnothing)=0</math>.<br />
<br />
(Case II.1) <math>\text{Core}=\varnothing</math>. Then either (II.1.1) there exist two nonintersecting subsets A and B, <math>C(A)=C(B)=1</math>, but f<math>(A)f(B)=0</math>, a contradiction, or (II.1.2) all subsets has <math>C(T)=0</math>, which is easily confirmed to satisfy the condition <math>f(T_1)f(T_2)=f(T_1 \cap T_2)f(T_1 \cup T_2)</math>. There is one coloring in this case.<br />
<br />
(Case II.2) Core = a subset of 1 element. WLOG, C(S_1)=1. Then <math>f(S_1)=1</math>, and subsets containing element 1 may be colored Blue. <math>f(S_1 \cup S_m)f(S_1\cup S_n)=f(S_1 \cup S_m \cup S_n)</math>, namely <math>C(S_1 \cup S_m \cup S_n)=C(S_m \cup S_1)C(S_n \cup S_1)</math>. Now S_1 functions as the <math>\varnothing</math> in case I, with <math>n-1</math> elements to combine into a base of <math>n-1</math> two-element sets, and all the other subsets are determined. There are <math>2^{n-1}</math> colorings for each choice of core. However, there are nC1 = n such cores. Hence altogether there are <math>n2^{n-1}</math> colorings in this case.<br />
<br />
(Case II.3) Core = a subset of 2 elements. WLOG, let <math>C(S_1 \cup S_2)=1</math>. Only subsets containing the core may be colored blue. With the same reasoning as in the preceding case, there are <math>(nC2)2^{n-2}</math> colorings.<br />
<br />
<math>\dots</math><br />
<br />
(Case II.n+1) Core = S. Then <math>C(S)=1</math>, with all other subsets <math>C(T)=0</math>, there is <math>1=\dbinom{n}{n}2^0</math><br />
<br />
Combining all the cases, we have <math>1+\left[2^n+\dbinom{n}{1}2^{n-1}+\dbinom{n}{2}2^{n-2}+ \cdots + \dbinom{n}{n}2^0\right]=\boxed{1+3^n}</math> colorings.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1309622013 USAMO Problems/Problem 12020-08-07T09:21:18Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We will use construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
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.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,<br />
<cmath>\begin{align*}<br />
\angle BRY'+\angle BPY'<br />
&=180^\circ-\theta+\theta\\<br />
&=180^\circ.<br />
\end{align*}</cmath><br />
<br />
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.<br />
<br />
Since points <math>Y'</math> and <math>Z'</math> lie on <math>AP,</math> we know that,<br />
<cmath>Y'=\omega_B\cap AP</cmath><br />
and that<br />
<cmath>Z'=\omega_C\cap AP.</cmath><br />
<br />
Hence, the points <math>Y'</math> and <math>Z'</math> coincide with the given points <math>Y</math> and <math>Z,</math> respectively.<br />
<br />
Since quadrilateral <math>AQTR</math> is also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle Y'TZ'<br />
&=180^\circ-\angle RTQ\\<br />
&=180^\circ-(180^\circ-\angle RAQ)\\<br />
&=\angle RAQ\\<br />
&=\alpha.<br />
\end{align*}</cmath><br />
<br />
Similarly, since quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle TY'Z'<br />
&=180^\circ-\angle RY'P\\<br />
&=180^\circ-(180^\circ-\angle RBP)\\<br />
&=\angle RBP\\<br />
&=\beta,<br />
\end{align*}</cmath><br />
and,<br />
<cmath>\begin{align*}<br />
\angle Y'Z'T<br />
&=180^\circ-\angle PZ'Q\\<br />
&=180^\circ-(180^\circ-\angle PCQ)\\<br />
&=\angle PCQ\\<br />
&=\gamma.<br />
\end{align*}</cmath><br />
<br />
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><br />
<br />
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.<br />
<br />
Hence, quadrilateral <math>AQTX</math> is cyclic.<br />
<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564));<br />
draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
Since the angles <math>\angle ART</math> and <math>\angle AXT</math> are inscribed in the same arc <math>\overarc{AT},</math> we have,<br />
<cmath>\begin{align*}<br />
\angle AXT<br />
&=\angle ART\\<br />
&=\theta.<br />
\end{align*}</cmath><br />
<br />
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,<br />
<cmath>Y'X/XZ'=BP/PC.</cmath><br />
<br />
Since <math>Y'=Y</math> and <math>Z'=Z</math> hence,<br />
<cmath>YX/XZ=BP/PC,</cmath><br />
<br />
as required.<br />
<br />
Proof by Negia<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1309612013 USAMO Problems/Problem 12020-08-07T09:20:42Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We will use construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
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.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,<br />
<cmath>\begin{align*}<br />
\angle BRY'+\angle BPY'<br />
&=180^\circ-\theta+\theta\\<br />
&=180^\circ.<br />
\end{align*}</cmath><br />
<br />
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.<br />
<br />
Since points <math>Y'</math> and <math>Z'</math> lie on <math>AP,</math> we know that,<br />
<cmath>Y'=\omega_B\cap AP</cmath><br />
and that<br />
<cmath>Z'=\omega_C\cap AP.</cmath><br />
<br />
Hence, the points <math>Y'</math> and <math>Z'</math> coincide with the given points <math>Y</math> and <math>Z,</math> respectively.<br />
<br />
Since quadrilateral <math>AQTR</math> is also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle Y'TZ'<br />
&=180^\circ-\angle RTQ\\<br />
&=180^\circ-(180^\circ-\angle RAQ)\\<br />
&=\angle RAQ\\<br />
&=\alpha.<br />
\end{align*}</cmath><br />
<br />
Similarly, since quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle TY'Z'<br />
&=180^\circ-\angle RY'P\\<br />
&=180^\circ-(180^\circ-\angle RBP)\\<br />
&=\angle RBP\\<br />
&=\beta,<br />
\end{align*}</cmath><br />
and,<br />
<cmath>\begin{align*}<br />
\angle Y'Z'T<br />
&=180^\circ-\angle PZ'Q\\<br />
&=180^\circ-(180^\circ-\angle PCQ)\\<br />
&=\angle PCQ\\<br />
&=\gamma.<br />
\end{align*}</cmath><br />
<br />
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><br />
<br />
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.<br />
<br />
Hence, quadrilateral <math>AQTX</math> is cyclic.<br />
<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564));<br />
draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
Since the angles <math>\angle ART</math> and <math>\angle AXT</math> are inscribed in the same arc <math>\overarc{AT},</math> we have,<br />
<cmath>\begin{align*}<br />
\angle AXT<br />
&=\angle ART\\<br />
&=\theta.<br />
\end{align*}</cmath><br />
<br />
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,<br />
<cmath>Y'X/XZ'=BP/PC.</cmath><br />
<br />
Since <math>Y'=Y</math> and <math>Z'=Z</math> hence,<br />
<cmath>YX/XZ=BP/PC,</cmath><br />
<br />
as required.<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1307892013 USAMO Problems/Problem 12020-08-06T16:28:55Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We can use some construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
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.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,<br />
<cmath>\begin{align*}<br />
\angle BRY'+\angle BPY'<br />
&=180^\circ-\theta+\theta\\<br />
&=180^\circ.<br />
\end{align*}</cmath><br />
<br />
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.<br />
<br />
Since points <math>Y'</math> and <math>Z'</math> lie on <math>AP,</math> we know that,<br />
<cmath>Y'=\omega_B\cap AP</cmath><br />
and that<br />
<cmath>Z'=\omega_C\cap AP.</cmath><br />
<br />
Hence, the points <math>Y'</math> and <math>Z'</math> coincide with the given points <math>Y</math> and <math>Z,</math> respectively.<br />
<br />
Since quadrilateral <math>AQTR</math> is also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle Y'TZ'<br />
&=180^\circ-\angle RTQ\\<br />
&=180^\circ-(180^\circ-\angle RAQ)\\<br />
&=\angle RAQ\\<br />
&=\alpha.<br />
\end{align*}</cmath><br />
<br />
Similarly, since quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle TY'Z'<br />
&=180^\circ-\angle RY'P\\<br />
&=180^\circ-(180^\circ-\angle RBP)\\<br />
&=\angle RBP\\<br />
&=\beta,<br />
\end{align*}</cmath><br />
and,<br />
<cmath>\begin{align*}<br />
\angle Y'Z'T<br />
&=180^\circ-\angle PZ'Q\\<br />
&=180^\circ-(180^\circ-\angle PCQ)\\<br />
&=\angle PCQ\\<br />
&=\gamma.<br />
\end{align*}</cmath><br />
<br />
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><br />
<br />
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.<br />
<br />
Hence, quadrilateral <math>AQTX</math> is cyclic.<br />
<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-0.10264330299819162,1.125351256231488));<br />
draw((-3.3284001481939356,0.7057864725120093)--(-5.053354907372894,2.4694710603912564));<br />
draw((-3.6988888888888977,6.426666666666669)--(-0.10264330299819162,1.125351256231488));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.6988888888888977,6.426666666666669), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.61,-5), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.09,-5), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.958888888888898,-5), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.5968131669050584,1.8770271258031248), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-5.053354907372894,2.4694710603912564), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.143912404905382,-2.142970212141873), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.413789986031826,2.0243286531799747), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X$", (-3.3284001481939356,0.7057864725120093), dir(220) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.7922953932137468,0.6108747864253139), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.8024625203461,-5), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.10264330299819162,1.125351256231488), dir(275) * labelscalefactor); <br />
</asy><br />
<br />
Since the angles <math>\angle ART</math> and <math>\angle AXT</math> are inscribed in the same arc <math>\overarc{AT},</math> we have,<br />
<cmath>\begin{align*}<br />
\angle AXT<br />
&=\angle ART\\<br />
&=\theta.<br />
\end{align*}</cmath><br />
<br />
Consider <math>\triangle TY'X</math> and <math>\triangle ABP.</math> We know that <math>\angle XY'T</math><br />
<br />
(Solution in progress)<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1307702013 USAMO Problems/Problem 12020-08-06T14:42:45Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We can use some construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
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.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.65,6.49333333333334), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.556666666666688,-4.866666666666666), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.15,-4.866666666666666), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.9033333333333546,-4.866666666666666), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.6566666666666455,2.0133333333333376), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-4.996666666666688,2.6), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.09,-2.04), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.3566666666666882,2.1333333333333377), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X'$", (-3.276666666666688,0.8133333333333368), dir(20) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.8433333333333121,0.72), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.7433333333333545,-4.8933333333333335), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.05,1.2266666666666703), dir(105) * labelscalefactor); <br />
</asy><br />
<br />
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,<br />
<cmath>\begin{align*}<br />
\angle BRY'+\angle BPY'<br />
&=180^\circ-\theta+\theta\\<br />
&=180^\circ.<br />
\end{align*}</cmath><br />
<br />
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.<br />
<br />
Since points <math>Y'</math> and <math>Z'</math> lie on <math>AP,</math> we know that,<br />
<cmath>Y'=\omega_B\cap AP</cmath><br />
and that<br />
<cmath>Z'=\omega_C\cap AP.</cmath><br />
<br />
Hence, the points <math>Y'</math> and <math>Z'</math> coincide with the given points <math>Y</math> and <math>Z,</math> respectively.<br />
<br />
Since quadrilateral <math>AQTR</math> is also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle Y'TZ'<br />
&=180^\circ-\angle RTQ\\<br />
&=180^\circ-(180^\circ-\angle RAQ)\\<br />
&=\angle RAQ\\<br />
&=\alpha.<br />
\end{align*}</cmath><br />
<br />
Similarly, since quadrilaterals <math>CQZ'P,</math> and <math>AQTR</math> are also cyclic, we have,<br />
<cmath>\begin{align*}<br />
\angle Y'TZ'<br />
&=180^\circ-\angle RTQ\\<br />
&=180^\circ-(180^\circ-\angle RAQ)\\<br />
&=\angle RAQ\\<br />
&=\alpha.<br />
\end{align*}</cmath><br />
(Solution in progress)<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2016_AMC_10A_Problems/Problem_4&diff=1306702016 AMC 10A Problems/Problem 42020-08-05T16:50:31Z<p>Negia: /* Solution */</p>
<hr />
<div>==Problem==<br />
<br />
The remainder can be defined for all real numbers <math>x</math> and <math>y</math> with <math>y \neq 0</math> by <cmath>\text{rem} (x ,y)=x-y\left \lfloor \frac{x}{y} \right \rfloor</cmath>where <math>\left \lfloor \tfrac{x}{y} \right \rfloor</math> denotes the greatest integer less than or equal to <math>\tfrac{x}{y}</math>. What is the value of <math>\text{rem} (\tfrac{3}{8}, -\tfrac{2}{5} )</math>?<br />
<br />
<math>\textbf{(A) } -\frac{3}{8} \qquad \textbf{(B) } -\frac{1}{40} \qquad \textbf{(C) } 0 \qquad \textbf{(D) } \frac{3}{8} \qquad \textbf{(E) } \frac{31}{40}</math><br />
<br />
==Solution==<br />
<br />
The value, by definition, is <cmath>\begin{align*}<br />
\text{rem}\left(\frac{3}{8},-\frac{2}{5}\right)<br />
&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{\frac{3}{8}}{-\frac{2}{5}}\right\rfloor \\<br />
&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{3}{8}\times\frac{-5}{2}\right\rfloor \\<br />
&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left\lfloor\frac{-15}{16}\right\rfloor\\<br />
&= \frac{3}{8}-\left(-\frac{2}{5}\right)\left(-1\right)\\<br />
&= \frac{3}{8}-\frac{2}{5}\\<br />
&= \boxed{\textbf{(B) } -\frac{1}{40}}.<br />
\end{align*}</cmath><br />
<br />
==Solution 2==<br />
Do note that the denominator of the answer will be a multiple of 5 and 8 (40) and that the answer will also be negative. The only answer choice that satisfies this is <math>\boxed{B}</math><br />
<br />
==Video Solution==<br />
https://youtu.be/VIt6LnkV4_w?t=195<br />
<br />
~IceMatrix<br />
<br />
https://youtu.be/CrW0Yx5tqV0<br />
<br />
~savannahsolver<br />
<br />
==See Also==<br />
{{AMC10 box|year=2016|ab=A|num-b=3|num-a=5}}<br />
{{AMC12 box|year=2016|ab=A|num-b=2|num-a=4}}<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1306692013 USAMO Problems/Problem 12020-08-05T16:49:42Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We can use some construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
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.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.65,6.49333333333334), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.556666666666688,-4.866666666666666), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.15,-4.866666666666666), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.9033333333333546,-4.866666666666666), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.6566666666666455,2.0133333333333376), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-4.996666666666688,2.6), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.09,-2.04), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.3566666666666882,2.1333333333333377), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X'$", (-3.276666666666688,0.8133333333333368), dir(20) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.8433333333333121,0.72), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.7433333333333545,-4.8933333333333335), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.05,1.2266666666666703), dir(105) * labelscalefactor); <br />
</asy><br />
<br />
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,<br />
<cmath>\begin{align*}<br />
\angle BRY'+\angle BPY'<br />
&=180^\circ-\theta+\theta\\<br />
&=180^\circ.<br />
\end{align*}</cmath><br />
<br />
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 quadrilateral <math>CQZ'P</math> is also cyclic.<br />
<br />
(Solution in progress)<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1306682013 USAMO Problems/Problem 12020-08-05T16:49:15Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We can use some construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
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.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.65,6.49333333333334), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.556666666666688,-4.866666666666666), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.15,-4.866666666666666), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.9033333333333546,-4.866666666666666), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.6566666666666455,2.0133333333333376), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-4.996666666666688,2.6), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.09,-2.04), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.3566666666666882,2.1333333333333377), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X'$", (-3.276666666666688,0.8133333333333368), dir(20) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.8433333333333121,0.72), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.7433333333333545,-4.8933333333333335), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.05,1.2266666666666703), dir(105) * labelscalefactor); <br />
</asy><br />
<br />
We know that <math>\angle BRY'=180^\circ-\angle ARY'=180^\circ-\theta.</math> Hence, we have,<br />
<cmath>\begin{align*}<br />
\angle BRY'+\angle BPY'<br />
&=180^\circ-\theta+\theta\\<br />
&=180^\circ.<br />
\end{align*}</cmath><br />
<br />
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 quadrilateral <math>CQZ'P</math> is also cyclic.<br />
<br />
<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1306542013 USAMO Problems/Problem 12020-08-05T15:47:29Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We can use some construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
Now we let the intersections of <math>AP</math> with <math>QU</math> and <math>RV</math> be <math>Y'</math> and <math>Z',</math> respectively. This construction is as follows.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 1.9;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.65,6.49333333333334), N * labelscalefactor); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.556666666666688,-4.866666666666666), SW * labelscalefactor); <br />
dot((7.09,-5)); <br />
label("$C$", (7.15,-4.866666666666666), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.9033333333333546,-4.866666666666666), S * labelscalefactor); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.6566666666666455,2.0133333333333376), NE * labelscalefactor); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-4.996666666666688,2.6), dir(165) * labelscalefactor); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.09,-2.04), dir(170) * labelscalefactor); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.3566666666666882,2.1333333333333377), NE * labelscalefactor); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X'$", (-3.276666666666688,0.8133333333333368), dir(20) * labelscalefactor); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.8433333333333121,0.72), NE * labelscalefactor); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.7433333333333545,-4.8933333333333335), S * labelscalefactor); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.05,1.2266666666666703), dir(105) * labelscalefactor); <br />
</asy><br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1306532013 USAMO Problems/Problem 12020-08-05T13:47:35Z<p>Negia: /* Solution 4 */</p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We can use some construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
Now we let the intersections of <math>AP</math> with <math>QU</math> and <math>RV</math> be <math>Y'</math> and <math>Z',</math> respectively. This construction is as follows.<br />
<asy><br />
import graph; size(12cm); <br />
real labelscalefactor = 0.5;<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);<br />
pen dotstyle = black;<br />
pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); <br />
<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)--(7.09,-5)--cycle);<br />
draw((-3.6988888888888977,6.426666666666669)--(-7.61,-5)); <br />
draw((-7.61,-5)--(7.09,-5)); <br />
draw((7.09,-5)--(-3.6988888888888977,6.426666666666669)); <br />
draw((-3.6988888888888977,6.426666666666669)--(-2.958888888888898,-5)); <br />
draw((-5.053354907372894,2.4694710603912564)--(1.7922953932137468,0.6108747864253139)); <br />
draw((0.5968131669050584,1.8770271258031248)--(-5.8024625203461,-5));<br />
dot((-3.6988888888888977,6.426666666666669)); <br />
label("$A$", (-3.65,6.49333333333334), N); <br />
dot((-7.61,-5)); <br />
label("$B$", (-7.556666666666688,-4.866666666666666), SW); <br />
dot((7.09,-5)); <br />
label("$C$", (7.15,-4.866666666666666), SE * labelscalefactor); <br />
dot((-2.958888888888898,-5)); <br />
label("$P$", (-2.9033333333333546,-4.866666666666666), S); <br />
dot((0.5968131669050584,1.8770271258031248)); <br />
label("$Q$", (0.6566666666666455,2.0133333333333376), NE); <br />
dot((-5.053354907372894,2.4694710603912564)); <br />
label("$R$", (-4.996666666666688,2.6), NE); <br />
dot((-3.143912404905382,-2.142970212141873)); <br />
label("$Z'$", (-3.09,-2.04), NE); <br />
dot((-3.413789986031826,2.0243286531799747)); <br />
label("$Y'$", (-3.3566666666666882,2.1333333333333377), NE); <br />
dot((-3.3284001481939356,0.7057864725120093)); <br />
label("$X'$", (-3.276666666666688,0.8133333333333368), NE); <br />
dot((1.7922953932137468,0.6108747864253139)); <br />
label("$V$", (1.8433333333333121,0.72), NE); <br />
dot((-5.8024625203461,-5)); <br />
label("$U$", (-5.7433333333333545,-4.8933333333333335), NE); <br />
dot((-0.10264330299819162,1.125351256231488)); <br />
label("$T$", (-0.05,1.2266666666666703), NE); <br />
</asy><br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_1&diff=1305552013 USAMO Problems/Problem 12020-08-04T06:02:58Z<p>Negia: </p>
<hr />
<div>==Problem==<br />
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><br />
<br />
==Solution 1==<br />
<br />
<asy><br />
/* DRAGON 0.0.9.6<br />
Homemade Script by v_Enhance. */<br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
defaultpen(fontsize(10pt));<br />
/* Initialize Objects */<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
/* Draw objects */<br />
draw(A--B, rgb(0.6,0.6,0.0));<br />
draw(B--C, rgb(0.6,0.6,0.0));<br />
draw(C--A, rgb(0.6,0.6,0.0));<br />
draw(w_A, rgb(0.4,0.4,0.0));<br />
draw(w_B, rgb(0.4,0.4,0.0));<br />
draw(w_C, rgb(0.4,0.4,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(D--E, rgb(0.0,0.2,0.4));<br />
draw(P--D, rgb(0.0,0.2,0.4));<br />
draw(P--E, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
/* Place dots on each point */<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
/* Label points */<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));</asy><br />
<br />
In this solution, all lengths and angles are directed.<br />
<br />
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><br />
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.<br />
<br />
<br />
courtesy v_enhance<br />
----<br />
<br />
==Solution 2==<br />
<br />
[https://www.flickr.com/photos/127013945@N03/14800492500/lightbox/ Diagram]<br />
Refer to the Diagram link.<br />
<br />
By Miquel's Theorem, there exists a point at which <math>\omega_A, \omega_B, \omega_C</math> intersect. We denote this point by <math>M.</math> Now, we angle chase:<br />
<cmath>\angle YMX = 180^{\circ} - \angle YXM - \angle XYM</cmath><cmath>= 180^{\circ} - \angle AXM - \angle PYM</cmath><cmath>= \left(180^{\circ} - \angle ARM\right) - \angle PRM</cmath><cmath>= \angle BRM - \angle PRM</cmath><cmath>= \angle BRP = \angle BMP.</cmath><br />
In addition, we have<br />
<cmath>\angle ZMX = 180^{\circ} - \angle MZY - \angle ZYM - \angle YMX</cmath><cmath>= 180^{\circ} - \angle MZP - \angle PYM - \angle BMP</cmath><cmath>= 180^{\circ} - \angle MCP - \angle PBM - \angle BMP</cmath><cmath>= \left(180^{\circ} - \angle PBM - \angle BMP\right) - \angle MCP</cmath><cmath>= \angle BPM - \angle MCP</cmath><cmath>= 180^{\circ} - \angle MPC - \angle MCP</cmath><cmath>= \angle CMP.</cmath><br />
Now, by the Ratio Lemma, we have<br />
<cmath>\frac{XY}{XZ} = \frac{MY}{MZ} \cdot \frac{\sin \angle YMX}{\sin \angle ZMX}</cmath><cmath>= \frac{\sin \angle YZM}{\sin \angle ZYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MZY</math>)<cmath>= \frac{\sin \angle PZM}{\sin \angle PYM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{\sin \angle PCM}{\sin \angle PBM} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath><cmath>= \frac{MB}{MC} \cdot \frac{\sin \angle BMP}{\sin \angle CMP}</cmath> (by the Law of Sines in <math>\triangle MBC</math>)<cmath>= \frac{PB}{PC}</cmath> by the Ratio Lemma.<br />
The proof is complete.<br />
<br />
==Solution 3==<br />
Use directed angles modulo <math>\pi</math>.<br />
<br />
Lemma. <math>\angle{XRY} \equiv \angle{XQZ}.</math><br />
<br />
Proof. <cmath>\angle{XRY} \equiv \angle{XRA} - \angle{YRA} \equiv \angle{XQA} + \angle{YRB} \equiv \angle{XQA} + \angle{CPY} = \angle{XQA} + \angle{AQZ} = \angle{XQZ}.</cmath><br />
<br />
Now, it follows that (now not using directed angles)<br />
<cmath>\frac{XY}{YZ} = \frac{\frac{XY}{\sin \angle{XRY}}}{\frac{YZ}{\sin \angle{XQZ}}} = \frac{\frac{RY}{\sin \angle{RXY}}}{\frac{QZ}{\sin \angle{QXZ}}} = \frac{BP}{PC}</cmath><br />
using the facts that <math>ARY</math> and <math>APB</math>, <math>AQZ</math> and <math>APC</math> are similar triangles, and that <math>\frac{RA}{\sin \angle{RXA}} = \frac{QA}{\sin \angle{QXA}}</math> equals twice the circumradius of the circumcircle of <math>AQR</math>.<br />
<br />
==Solution 4==<br />
We can use some construction arguments to solve the problem.<br />
<br />
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><br />
<br />
<br />
Now we let the intersections of <math>AP</math> with <math>QU</math> and <math>RV</math> be <math>Y'</math> and <math>Z',</math> respectively. This construction is as follows.<br />
<asy><br />
import olympiad;<br />
import cse5;<br />
size(11cm);<br />
real lsf=0.8000;<br />
real lisf=2011.0;<br />
<br />
defaultpen(fontsize(10pt));<br />
<br />
pair A = (-1.0, 3.0);<br />
pair B = (-3.0, -3.0);<br />
pair C = (4.0, -3.0);<br />
pair P = (-0.6698198198198195, -3.0);<br />
pair Q = (1.1406465288818244, 0.43122416534181074);<br />
pair R = (-1.6269590345062048, 1.119122896481385);<br />
path w_A = circumcircle(A,Q,R);<br />
path w_B = circumcircle(B,P,R);<br />
path w_C = circumcircle(P,Q,C);<br />
pair O_A = midpoint(relpoint(w_A, 0)--relpoint(w_A, 0.5));<br />
pair O_B = midpoint(relpoint(w_B, 0)--relpoint(w_B, 0.5));<br />
pair O_C = midpoint(relpoint(w_C, 0)--relpoint(w_C, 0.5));<br />
pair X = (2)*(foot(O_A,A,P))-A;<br />
pair Y = (2)*(foot(O_B,A,P))-P;<br />
pair Z = (2)*(foot(O_C,A,P))-P;<br />
pair M = 2*foot(P,relpoint(O_B--O_C,0.5-10/lisf),relpoint(O_B--O_C,0.5+10/lisf))-P;<br />
pair D = (2)*(foot(O_B,X,M))-M;<br />
pair E = (2)*(foot(O_C,X,M))-M;<br />
<br />
draw(A--B--C--A, rgb(0.6,0.6,0.0));<br />
draw(A--P, rgb(0.0,0.2,0.4));<br />
draw(P--M, rgb(0.4,0.2,0.0));<br />
draw(R--M, rgb(0.4,0.2,0.0));<br />
draw(Q--M, rgb(0.4,0.2,0.0));<br />
draw(B--M, rgb(0.0,0.2,0.4));<br />
draw(C--M, rgb(0.0,0.2,0.4));<br />
draw(R--Y);<br />
draw((abs(dot(unit(M-P),unit(B-P))) < 1/2011) ? rightanglemark(M,P,B) : anglemark(M,P,B), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-R),unit(A-R))) < 1/2011) ? rightanglemark(M,R,A) : anglemark(M,R,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(M-X),unit(A-X))) < 1/2011) ? rightanglemark(M,X,A) : anglemark(M,X,A), rgb(0.0,0.8,0.8));<br />
draw((abs(dot(unit(D-X),unit(P-X))) < 1/2011) ? rightanglemark(D,X,P) : anglemark(D,X,P), rgb(0.0,0.8,0.8));<br />
<br />
dot(A);<br />
dot(B);<br />
dot(C);<br />
dot(P);<br />
dot(Q);<br />
dot(R);<br />
dot(X);<br />
dot(Y);<br />
dot(Z);<br />
dot(M);<br />
dot(D);<br />
dot(E);<br />
label("$A$", A, lsf * dir(110));<br />
label("$B$", B, lsf * unit(B-M));<br />
label("$C$", C, lsf * unit(C-M));<br />
label("$P$", P, lsf * unit(P-M) * 1.8);<br />
label("$Q$", Q, lsf * dir(90) * 1.6);<br />
label("$R$", R, lsf * unit(R-M) * 2);<br />
label("$X$", X, lsf * dir(-60) * 2);<br />
label("$Y$", Y, lsf * dir(45));<br />
label("$Z$", Z, lsf * dir(5));<br />
label("$M$", M, lsf * dir(M-P)*2);<br />
label("$D$", D, lsf * dir(150));<br />
label("$E$", E, lsf * dir(5));<br />
</asy><br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=1305252013 USAMO Problems/Problem 52020-08-04T05:05:21Z<p>Negia: </p>
<hr />
<div>Given postive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.<br />
{{MAA Notice}}<br />
<br />
==Solution==<br />
<br />
This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.<br />
<br />
<br />
Without Loss of Generality, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</math>.<br />
<br />
<br />
Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.<br />
<br />
<br />
Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another.<br />
<br />
<br />
This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_3&diff=1305182013 USAMO Problems/Problem 32020-08-04T04:50:50Z<p>Negia: </p>
<hr />
<div>Let <math>n</math> be a positive integer. There are <math>\tfrac{n(n+1)}{2}</math> marks, each with a black side and a white side, arranged into an equilateral triangle, with the biggest row containing <math>n</math> marks. Initially, each mark has the black side up. An ''operation'' is to choose a line parallel to the sides of the triangle, and flipping all the marks on that line. A configuration is called ''admissible'' if it can be obtained from the initial configuration by performing a finite number of operations. For each admissible configuration <math>C</math>, let <math>f(C)</math> denote the smallest number of operations required to obtain <math>C</math> from the initial configuration. Find the maximum value of <math>f(C)</math>, where <math>C</math> varies over all admissible configurations.<br />
<br />
<br />
<br />
{{MAA Notice}}</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=1305172013 USAMO Problems/Problem 52020-08-04T04:50:03Z<p>Negia: /* Solution */</p>
<hr />
<div>Given postive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.<br />
{{MAA Notice}}<br />
<br />
==Solution==<br />
<br />
This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.<br />
<br />
<br />
WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</math>.<br />
<br />
<br />
Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.<br />
<br />
<br />
Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another.<br />
<br />
<br />
This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=1305152013 USAMO Problems/Problem 52020-08-04T04:49:43Z<p>Negia: /* Solution */</p>
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<div>Given postive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.<br />
{{MAA Notice}}<br />
<br />
==Solution==<br />
<br />
This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.<br />
<br />
<br />
WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</math>.<br />
<br />
Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.<br />
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Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another.<br />
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This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_5&diff=1305142013 USAMO Problems/Problem 52020-08-04T04:48:43Z<p>Negia: /* Solution */</p>
<hr />
<div>Given postive integers <math>m</math> and <math>n</math>, prove that there is a positive integer <math>c</math> such that the numbers <math>cm</math> and <math>cn</math> have the same number of occurrences of each non-zero digit when written in base ten.<br />
{{MAA Notice}}<br />
<br />
==Solution==<br />
<br />
This solution is adopted from the [http://web.archive.org/web/20130606075948/http://amc.maa.org/usamo/2013/2013USAMO_Day1_Day2_Final_S.pdf official solution]. Both the problem and the solution were suggested by Richard Stong.<br />
<br />
<br />
WLOG, suppose <math>m \geq n \geq 1</math>. By prime factorization of <math>n</math>, we can find a positive integer <math>c_1</math> such that <math>c_1n=10^s n_1</math> where <math>n_1</math> is relatively prime to <math>10</math>. If a positive <math>k</math> is larger than <math>s</math>, then <math>(10^k c_1 m - c_1 n)= 10^s t</math>, where <math>t=10^{k-s} c_1m-n_1>0</math> is always relatively prime to <math>10</math>. Choose a <math>k</math> large enough so that <math>t</math> is larger than <math>c_1m</math>. We can find an integer <math>b\geq 1</math> such that <math>10^b-1</math> is divisible by <math>t</math>, and also larger than <math>10c_1m</math>. For example, let <math>b=\varphi(t)</math> and use Euler's theorem. Now, let <math>c_2=(10^b-1)/t</math>, and <math>c=c_1c_2</math>. We claim that <math>c</math> is the desired number.<br />
<br />
<br />
Indeed, since both <math>c_1m</math> and <math>n_1</math> are less than <math>t</math>, we see that the decimal expansion of both the fraction <math>(c_1m)/t = (cm)/(c_2t) = (cm)/(10^b-1)</math> and <math>n_1/t=(c_2n_1)/(10^b-1)</math> are repeated in <math>b</math>-digit. And we also see that <math> 10^k (c_1m)/t = (t+c_1n)/t= 1+10^s (n_1/t)</math>, therefore the two repeated <math>b</math>-digit expansions are cyclic shift of one another. This proves that <math>cm</math> and <math>c_2n_1</math> have the same number of occurrences of non-zero digits. Furthermore, <math>cn = c_2c_1n=10^s c_2n_1</math> also have the same number of occurrences of non-zero digits with <math>c_2n_1</math>.</div>Negiahttps://artofproblemsolving.com/wiki/index.php?title=2013_USAMO_Problems/Problem_6&diff=1305132013 USAMO Problems/Problem 62020-08-04T04:45:53Z<p>Negia: /* Solution */</p>
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<div>==Problem==<br />
Let <math>ABC</math> be a triangle. Find all points <math>P</math> on segment <math>BC</math> satisfying the following property: If <math>X</math> and <math>Y</math> are the intersections of line <math>PA</math> with the common external tangent lines of the circumcircles of triangles <math>PAB</math> and <math>PAC</math>, then <cmath>\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.</cmath><br />
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==Solution==<br />
Let circle <math>PAB</math> (i.e. the circumcircle of <math>PAB</math>), <math>PAC</math> be <math>\omega_1, \omega_2</math> with radii <math>r_1</math>, <math>r_2</math> and centers <math>O_1, O_2</math>, respectively, and <math>d</math> be the distance between their centers.<br />
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'''Lemma.''' <math>XY = \frac{r_1 + r_2}{d} \sqrt{d^2 - (r_1 - r_2)^2}.</math><br />
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Proof. Let the external tangent containing <math>X</math> meet <math>\omega_1</math> at <math>X_1</math> and <math>\omega_2</math> at <math>X_2</math>, and let the external tangent containing <math>Y</math> meet <math>\omega_1</math> at <math>Y_1</math> and <math>\omega_2</math> at <math>Y_2</math>. Then clearly <math>X_1 Y_1</math> and <math>X_2 Y_2</math> are parallel (for they are both perpendicular <math>O_1 O_2</math>), and so <math>X_1 Y_1 Y_2 X_2</math> is a trapezoid.<br />
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Now, <math>X_1 X^2 = XA \cdot XP = X_2 X^2</math> by Power of a Point, and so <math>X</math> is the midpoint of <math>X_1 X_2</math>. Similarly, <math>Y</math> is the midpoint of <math>Y_1 Y_2</math>. Hence, <math>XY = \frac{1}{2} (X_1 Y_1 + X_2 Y_2).</math> Let <math>X_1 Y_1</math>, <math>X_2 Y_2</math> meet <math>O_1 O_2</math> s at <math>Z_1, Z_2</math>, respectively. Then by similar triangles and the Pythagorean Theorem we deduce that <math>X_1 Z_1 = \frac{r_1 \sqrt{d^2 - (r_1 - r_2)^2}}{d}</math> and <math>\frac{r_2 \sqrt{d^2 - (r_1 - r_2)^2}}{d}</math>. But it is clear that <math>Z_1</math>, <math>Z_2</math> is the midpoint of <math>X_1 Y_1</math>, <math>X_2 Y_2</math>, respectively, so <math>XY = \frac{(r_1 + r_2)}{d} \sqrt{d^2 - (r_1 - r_2)^2},</math> as desired.<br />
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Lemma 2. Triangles <math>O_1 A O_2</math> and <math>BAC</math> are similar.<br />
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Proof. <math>\angle{AO_1 O_2} = \frac{\angle{PO_1 A}}{2} = \angle{ABC}</math> and similarly <math>\angle{AO_2 O_1} = \angle{ACB}</math>, so the triangles are similar by AA Similarity.<br />
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Also, let <math>O_1 O_2</math> intersect <math>AP</math> at <math>Z</math>. Then obviously <math>Z</math> is the midpoint of <math>AP</math> and <math>AZ</math> is an altitude of triangle <math>A O_1 O_2</math>.Thus, we can simplify our expression of <math>XY</math>:<br />
<cmath>XY = \frac{AB + AC}{BC} \cdot \frac{AP}{2 h_a} \sqrt{BC^2 - (AB - AC)^2},</cmath><br />
where <math>h_a</math> is the length of the altitude from <math>A</math> in triangle <math>ABC</math>. Hence, substituting into our condition and using <math>AB = c, BC = a, CA = b</math> gives<br />
<cmath>\left( \frac{2a h_a}{(b+c) \sqrt{a^2 - (b-c)^2}} \right)^2 + \frac{PB \cdot PC}{bc} = 1.</cmath><br />
Using <math>2 a h_a = 4[ABC] = \sqrt{(a + b + c)(a + b - c)(a - b + c)(-a + b + c)}</math> by Heron's Formula (where <math>[ABC]</math> is the area of triangle <math>ABC</math>, our condition becomes<br />
<cmath>\frac{(a + b + c)(-a + b + c)}{(b + c)^2} + \frac{PB \cdot PC}{bc} = 1,</cmath><br />
which by <math>(a + b + c)(-a + b + c) = (b + c)^2 - a^2</math> becomes<br />
<cmath>\frac{PB \cdot PC}{bc} = \frac{a^2 bc}{(b+c)^2}.</cmath><br />
Let <math>PB = x</math>; then <math>PC = a - x</math>. The quadratic in <math>x</math> is<br />
<cmath>x^2 - ax + \frac{a^2 bc}{(b+c)^2} = 0,</cmath><br />
which factors as<br />
<cmath>\left(x - \frac{ab}{b+c}\right)\left(x - \frac{ac}{b+c}\right) = 0.</cmath><br />
Hence, <math>PB = \frac{ab}{b+c}</math> or <math>\frac{ac}{b+c}</math>, and so the <math>P</math> corresponding to these lengths are our answer.<br />
{{MAA Notice}}</div>Negia