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Source: Iran TST 2008

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1275 posts

#1 h Jul 5, 2008, 10:10 PM • 2 Y

Y by Adventure10, Mango247

In the triangle $ABC$ , $\angle B$ is greater than $\angle C$ . $T$ is the midpoint of the arc $BAC$ from the circumcircle of $ABC$ and $I$ is the incenter of $ABC$ . $E$ is a point such that $\angle AEI=90^\circ$ and $AE\parallel BC$ . $TE$ intersects the circumcircle of $ABC$ for the second time in $P$ . If $\angle B=\angle IPB$ , find the angle $\angle A$ .

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#2 h Jul 6, 2008, 7:08 AM • 2 Y

Y by guptaamitu1, Adventure10

We have :
$\angle{CAE}+\angle {CAI}+\angle {AIE}=\pi$
From $AE||BE \rightarrow \angle{CAE}=\angle{ACB}$
Therefore $\angle{AIE}=\frac{\angle{B}-\angle{C}}{2}$
Other $\angle{APT}=\frac{\angle{B}-\angle{C}}{2}$
It gives $\angle{AIE}=\angle{APE}$ so $A,P,I,E$ lie on a circle .
Because $\angle{IEA}=\frac{\pi}{2}$ so $\angle{API}=\frac{\pi}{2}$
From condition $\angle{APB}+\angle{C}=\pi$ and $PBI=\angle{B}$ we have $\angle{A}=\frac{\pi}{2}$

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#3 h Sep 17, 2008, 12:26 PM • 1 Y

Y by Adventure10

a small error....the first eqn. RHS is equal to pi/2...

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#5 h Jun 30, 2019, 10:21 AM • 2 Y

Y by guptaamitu1, Adventure10

Iran TST 2008 P10 wrote:

In the triangle $ABC$ , $\angle B$ is greater than $\angle C$ . $T$ is the midpoint of the arc $BAC$ from the circumcircle of $ABC$ and $I$ is the incenter of $ABC$ . $E$ is a point such that $\angle AEI=90^\circ$ and $AE\parallel BC$ . $TE$ intersects the circumcircle of $ABC$ for the second time in $P$ . If $\angle B=\angle IPB$ , find the angle $\angle A$ .

Solution: Let $\odot (AIE)$ $\cap$ $\odot (ABC)$ $=$ $D$ . Let $AA'||BC$ with $A' \in \odot (ABC)$ .
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.831784745493193, xmax = 17.492591423028003, ymin = -10.676873790658856, ymax = 6.749497293093495; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-4.06,4.19)--(-6.6,-5.07)--(4.68,-4.99)--cycle, linewidth(1.2) + rvwvcq); /* draw figures */ draw((-4.06,4.19)--(-6.6,-5.07), linewidth(1.2) + rvwvcq); draw((-6.6,-5.07)--(4.68,-4.99), linewidth(1.2) + rvwvcq); draw((4.68,-4.99)--(-4.06,4.19), linewidth(1.2) + rvwvcq); draw(circle((-0.9840987783166554,-1.6320722573515867), 6.5846559283971775), linewidth(0.8) + linetype("4 4")); draw((-4.06,4.19)--(-0.9374002653172535,-8.216562590267243), linewidth(0.4)); draw((-6.6,-5.07)--(3.784841528296612,2.9082913896854476), linewidth(0.4)); draw((-7.334197787673028,0.10974755730987962)--(4.68,-4.99), linewidth(0.4)); draw(shift((-3.2892794840301782,1.1278104357156291))*xscale(3.1576914100888476)*yscale(3.1576914100888476)*arc((0,0),1,-75.872693601157,104.127306398843), linewidth(0.4)); draw(shift((-3.2892794840301782,1.1278104357156291))*xscale(3.1576914100888476)*yscale(3.1576914100888476)*arc((0,0),1,104.12730639884302,284.127306398843), linewidth(0.4)); draw((-4.06,4.19)--(2.00891459611709,4.233041947490193), linewidth(0.4)); draw((-6.165234649953677,2.4316113886784727)--(2.091802443366688,-7.454144514703174), linewidth(0.4)); draw((2.00891459611709,4.233041947490193)--(2.091802443366688,-7.454144514703174), linewidth(0.4)); draw((-2.5620696258493183,4.200623619674828)--(-2.518558968060357,-1.9343791285687424), linewidth(0.4)); /* dots and labels */ dot((-4.06,4.19),dotstyle); label("$A$", (-4.452617955545645,4.562187267408717), NE * labelscalefactor); dot((-6.6,-5.07),dotstyle); label("$B$", (-7.312946450671894,-5.364835156852966), NE * labelscalefactor); dot((4.68,-4.99),dotstyle); label("$C$", (5.161931607903928,-5.196580539492599), NE * labelscalefactor); dot((-0.9374002653172535,-8.216562590267243),dotstyle); label("$M_A$", (-1.5682530865107733,-9.042400364872428), NE * labelscalefactor); dot((-1.0307972913160572,4.95241807556407),dotstyle); label("$M_{BC}$", (-1.1115619822469187,5.547678597662299), NE * labelscalefactor); dot((2.00891459611709,4.233041947490193),dotstyle); label("$A'$", (2.4217849823207995,4.682369136951837), NE * labelscalefactor); dot((3.784841528296612,2.9082913896854476),dotstyle); label("$M_B$", (4.224513025467594,3.264223076343025), NE * labelscalefactor); dot((-7.334197787673028,0.10974755730987962),dotstyle); label("$M_C$", (-8.418619650468594,0.35582183339952933), NE * labelscalefactor); dot((-2.518558968060357,-1.9343791285687424),dotstyle); label("$I$", (-2.4335625472212348,-1.687269948833505), NE * labelscalefactor); dot((-6.165234649953677,2.4316113886784727),dotstyle); label("$D$", (-6.808182598590791,2.7834955981705467), NE * labelscalefactor); dot((-2.5620696258493183,4.200623619674828),dotstyle); label("$E$", (-2.1931988081349956,3.360368571977521), NE * labelscalefactor); dot((2.091802443366688,-7.454144514703174),dotstyle); label("$F$", (2.6140759735897907,-7.840581669441232), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Applying Reim's Theorem we have, if $DI$ $\cap$ $\odot (ABC)$ $=$ $F$ $\implies$ $EI$ $||$ $A'F$ $\implies$ $F$ is $A-$ antipode and Applying Reim's Theorem again $\implies$ $EI$ $||$ $M_{BC}M_A$ $\implies$ $D \equiv P$ . Now, $\angle IPB$ $=$ $90^{\circ}$ $=$ $\angle B$ $\implies$ $\angle A=90^{\circ}$

This post has been edited 1 time. Last edited by AlastorMoody, Jun 30, 2019, 10:26 AM

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#6 h Jun 30, 2019, 2:27 PM • 1 Y

Y by Adventure10

We can assume that triangle $ABC$ is inscribed in the unit circle on Argand Plane. Let $a,b,c\in\mathbb C$ be numbers, such that $a^2,b^2,c^2$ are vertices of our triangle (so $|a|=|b|=|c|=1$ )and it's incenter is $-(ab+bc+ca)$ . Then $t=bc$ .
$AE||BC\implies\frac{a^2-e}{b^2-c^2}=\overline{\left(\frac{a^2-e}{b^2-c^2}\right)}\implies \overline e=\frac{a^2-e}{b^2c^2}+\frac{1}{a^2}$ Thus
$IE\perp AE\implies\frac{a^2-e}{e+ab+bc+ca}=\overline{\left(\frac{a^2-e}{e+ab+bc+ca}\right)}=\frac{a^2(a^2-e)}{a^2(a^2-e)+b^2c^2+abc(a+b+c)}$ it means
$e=\frac{a^4+b^2c^2+abc(b+c)-a^3(b+c)}{2a^2}$ .
$P$ lies on the circumcircle of triangle $ABC$ so $|p|=1$
$\frac{p-t}{e-t}=\frac{2a^2(p-bc)}{(a^2-bc)(a^2-ab-bc-ca)}$ Points $P,T,E$ lie on one line:
$\frac{p-t}{e-t}=\overline{\left(\frac{p-t}{e-t}\right)}\iff$ $\frac{2a^2(p-bc)}{(a^2-bc)(a^2-ab-bc-ca)}=\frac{2a^2bc(p-bc)}{p(a^2-bc)(bc-a^2-ab-ac)}\iff$ $p=\frac{a^2-bc-ab-ac}{bc-a^2-ab-ac}\cdot bc$ $\angle B=\angle IPB$ and $\angle B$ is greater than $\angle C$ so we have directed angles equality:
$\frac{\frac{p+ab+bc+ca}{p-b^2}}{|\frac{p+ab+bc+ca}{p-b^2}|}=\frac{\frac{a^2-b^2}{c^2-b^2}}{|\frac{a^2-b^2}{c^2-b^2}|}$ square both sides and use the identity $|x|^2=x\cdot \overline x$ to get
$\frac{\frac{p+ab+bc+ca}{p-b^2}}{\overline{\left(\frac{p+ab+bc+ca}{p-b^2}\right)}}=\frac{\frac{a^2-b^2}{c^2-b^2}}{\overline{\left(\frac{a^2-b^2}{c^2-b^2}\right)}}$ Auxiliary calculations:
$\frac{p+ab+bc+ca}{p-b^2}=\frac{a(abc-(a+b+c)(ab+bc+ca))}{b(b+c)(a+b)(a-c)}$ $\overline{\left(\frac{p+ab+bc+ca}{p-b^2}\right)}=\frac{b(abc-(a+b+c)(ab+bc+ca))}{a(b+c)(a+b)(-a+c)}$ so LHS is equal to $\frac{-a^2}{b^2}$
Again AUX
$\overline{\left(\frac{a^2-b^2}{c^2-b^2}\right)}=\frac{a^2-b^2}{c^2-b^2}\cdot\frac{c^2}{a^2}$ so RHS is equal to $\frac{a^2}{c^2}$ .
We gained $\frac{-a^2}{b^2}=\frac{a^2}{c^2}$ but $|a|=1\implies a\neq 0$ hence
$b^2+c^2=0$ which means that circumcenter is the midpoint of $BC$ .

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#7 h Jun 30, 2019, 2:36 PM • 1 Y

Y by Adventure10

AlastorMoody wrote:

Iran TST 2008 P10 wrote:

In the triangle $ABC$ , $\angle B$ is greater than $\angle C$ . $T$ is the midpoint of the arc $BAC$ from the circumcircle of $ABC$ and $I$ is the incenter of $ABC$ . $E$ is a point such that $\angle AEI=90^\circ$ and $AE\parallel BC$ . $TE$ intersects the circumcircle of $ABC$ for the second time in $P$ . If $\angle B=\angle IPB$ , find the angle $\angle A$ .

Solution: Let $\odot (AIE)$ $\cap$ $\odot (ABC)$ $=$ $D$ . Let $AA'||BC$ with $A' \in \odot (ABC)$ .
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.831784745493193, xmax = 17.492591423028003, ymin = -10.676873790658856, ymax = 6.749497293093495; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); draw((-4.06,4.19)--(-6.6,-5.07)--(4.68,-4.99)--cycle, linewidth(1.2) + rvwvcq); /* draw figures */ draw((-4.06,4.19)--(-6.6,-5.07), linewidth(1.2) + rvwvcq); draw((-6.6,-5.07)--(4.68,-4.99), linewidth(1.2) + rvwvcq); draw((4.68,-4.99)--(-4.06,4.19), linewidth(1.2) + rvwvcq); draw(circle((-0.9840987783166554,-1.6320722573515867), 6.5846559283971775), linewidth(0.8) + linetype("4 4")); draw((-4.06,4.19)--(-0.9374002653172535,-8.216562590267243), linewidth(0.4)); draw((-6.6,-5.07)--(3.784841528296612,2.9082913896854476), linewidth(0.4)); draw((-7.334197787673028,0.10974755730987962)--(4.68,-4.99), linewidth(0.4)); draw(shift((-3.2892794840301782,1.1278104357156291))*xscale(3.1576914100888476)*yscale(3.1576914100888476)*arc((0,0),1,-75.872693601157,104.127306398843), linewidth(0.4)); draw(shift((-3.2892794840301782,1.1278104357156291))*xscale(3.1576914100888476)*yscale(3.1576914100888476)*arc((0,0),1,104.12730639884302,284.127306398843), linewidth(0.4)); draw((-4.06,4.19)--(2.00891459611709,4.233041947490193), linewidth(0.4)); draw((-6.165234649953677,2.4316113886784727)--(2.091802443366688,-7.454144514703174), linewidth(0.4)); draw((2.00891459611709,4.233041947490193)--(2.091802443366688,-7.454144514703174), linewidth(0.4)); draw((-2.5620696258493183,4.200623619674828)--(-2.518558968060357,-1.9343791285687424), linewidth(0.4)); /* dots and labels */ dot((-4.06,4.19),dotstyle); label("$A$", (-4.452617955545645,4.562187267408717), NE * labelscalefactor); dot((-6.6,-5.07),dotstyle); label("$B$", (-7.312946450671894,-5.364835156852966), NE * labelscalefactor); dot((4.68,-4.99),dotstyle); label("$C$", (5.161931607903928,-5.196580539492599), NE * labelscalefactor); dot((-0.9374002653172535,-8.216562590267243),dotstyle); label("$M_A$", (-1.5682530865107733,-9.042400364872428), NE * labelscalefactor); dot((-1.0307972913160572,4.95241807556407),dotstyle); label("$M_{BC}$", (-1.1115619822469187,5.547678597662299), NE * labelscalefactor); dot((2.00891459611709,4.233041947490193),dotstyle); label("$A'$", (2.4217849823207995,4.682369136951837), NE * labelscalefactor); dot((3.784841528296612,2.9082913896854476),dotstyle); label("$M_B$", (4.224513025467594,3.264223076343025), NE * labelscalefactor); dot((-7.334197787673028,0.10974755730987962),dotstyle); label("$M_C$", (-8.418619650468594,0.35582183339952933), NE * labelscalefactor); dot((-2.518558968060357,-1.9343791285687424),dotstyle); label("$I$", (-2.4335625472212348,-1.687269948833505), NE * labelscalefactor); dot((-6.165234649953677,2.4316113886784727),dotstyle); label("$D$", (-6.808182598590791,2.7834955981705467), NE * labelscalefactor); dot((-2.5620696258493183,4.200623619674828),dotstyle); label("$E$", (-2.1931988081349956,3.360368571977521), NE * labelscalefactor); dot((2.091802443366688,-7.454144514703174),dotstyle); label("$F$", (2.6140759735897907,-7.840581669441232), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Applying Reim's Theorem we have, if $DI$ $\cap$ $\odot (ABC)$ $=$ $F$ $\implies$ $EI$ $||$ $A'F$ $\implies$ $F$ is $A-$ antipode and Applying Reim's Theorem again $\implies$ $EI$ $||$ $M_{BC}M_A$ $\implies$ $D \equiv P$ . Now, $\angle IPB$ $=$ $90^{\circ}$ $=$ $\angle B$ $\implies$ $\angle A=90^{\circ}$

How do you import a graph from Geogebra?

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#8 h Sep 13, 2022, 11:40 AM

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JustKeepRunning wrote:

How do you import a graph from Geogebra?

It can be exported from GeoGebra Classic. If you find it easier to draw figures on GeoGebra Geometry, then you can first save it, and reopen in GeoGebra Classic.

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