angle BAS = angle CAM

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angle BAS = angle CAM

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Source: All-Russian Olympiad 2010 grade 10 P-3

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Ovchinnikov Denis

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Ovchinnikov Denis

#1 h Sep 9, 2010, 1:29 PM • 3 Y

Y by HWenslawski, Adventure10, Mango247

Let $O$ be the circumcentre of the acute non-isosceles triangle $ABC$ . Let $P$ and $Q$ be points on the altitude $AD$ such that $OP$ and $OQ$ are perpendicular to $AB$ and $AC$ respectively. Let $M$ be the midpoint of $BC$ and $S$ be the circumcentre of triangle $OPQ$ . Prove that $\angle BAS =\angle CAM$ .

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#2 h Sep 10, 2010, 7:12 AM • 4 Y

Y by NHN, Everything999, Adventure10, and 1 other user

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kyyuanmathcount

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kyyuanmathcount

#3 h Dec 13, 2010, 5:18 PM • 4 Y

Y by Adventure10, Mango247, and 2 other users

Easily with angles, we see that OPQ is similar to ABC in that orientation --> Let perpendicular from S to AD be ST.

Note from simple angles that <SOA = 90, so AOST is cyclic and <SAT = <SOT = <OAM.

It follows that <BAS = <BAD + <SAT = <OAC + <OAM = <CAM.

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#4 h Oct 15, 2011, 1:28 PM • 2 Y

Y by Adventure10, Mango247

oh its really nice problem but....!
suppose $op$ intersect the $AB$ at $Y$ and $OQ$ intersect $AC$ at $x$ .
it is really easy to show that $XY$ is parallel to $BC$ so if midpoint of $XY$ is $N$ we know that $XAN=CAM$
lemma:
in the triangle $AXY$ if the tangent to the circumcircle of it on $X$ and $Y$ intersect each other on $Z$ than $AZ$ concides with a symmedian of $AXY$ ..!
proof lemma:
on attachment!!
so it easy to show that $S$ lie on $AZ$ ..!(not so easy

if you need i will say it

)

Attachments:

lemma.pdf (196kb)

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112 posts

#5 h Feb 2, 2013, 7:36 PM • 1 Y

Y by Adventure10

It is enough if we prove that $\angle DAM=\angle OAS$ SO we prove that the triangles $DAM$ and $AOS$ are similar .
for proving this first we prove $\angle AOS=\angle ADM=90^{\circ}$
then it is easy to prove that $\frac{AD}{DM}=\frac{AO}{OS}$ by using this:
$DM=\frac{AC^2-AB^2}{2BC}$

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#6 h Jan 27, 2015, 4:08 PM • 2 Y

Y by Adventure10, Mango247

Let the tangent at $A$ to the circumcircle of $ABC$ intersects $BC$ at $D$ .Now,by an easy angle chase and using that $MOAD$ is cyclic,it is enough to prove that $DBO$ is similar to $APS$ and we have that angles $DBO$ and $APS$ are equal,so it is enough to prove that $AP/PS=BD/OB$ but this is easy cause $AOP$ is similar to $ADB$ and $OAB$ is similar to $OPS$ ,so done.

This post has been edited 1 time. Last edited by junioragd, Jan 27, 2015, 5:36 PM

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#7 h Jan 27, 2015, 5:35 PM • 3 Y

Y by amar_04, Adventure10, Mango247

My solution:

Let $N$ be the projection of $O$ on $PQ$ .

Since $OP \perp AB, OQ \perp AC, PQ \perp BC$ ,
so we get $\triangle OPQ \cup S \cup N \sim \triangle ABC \cup O \cup D$ and $\angle AOS=90^{\circ}$ ,
hence from $AO:OS=AD:ON=AD:DM \Longrightarrow Rt \triangle AOS \sim Rt \triangle ADM$ ,
so we get $\angle DAM=\angle OAS \Longrightarrow \angle BAS=\angle CAM$ .

Q.E.D

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#8 h Dec 12, 2015, 10:00 PM • 1 Y

Y by Adventure10

WLOG let $AB<AC$ , so that $O$ is in the interior of $\triangle ADC$ . Let $M_C=OP\cap AB$ and $M_B=OQ\cap AC$ . Note that by angle-chasing $\angle QPO=\angle APM_C=\angle ABD$ and $\angle PQO\equiv\angle AQM_B=\angle ACD$ , so $\triangle ABC\sim\triangle OPQ$ . Hence there exists a spiral similarity sending $\triangle ABC$ to $\triangle OPQ$ . Note that this sends $PQ$ to $BC$ , so the angle of rotation of this spiral similarity is $90^\circ$ . Thus, since $A$ is sent to $O$ and $O$ is sent to $S$ , we have $AO\perp OS$ .

Now let $X$ be the orthogonal projection of $O$ onto $PQ\equiv AD$ . Remark that by similarity and the fact that $MOXD$ is a rectangle, $\dfrac{MD}{DA}=\dfrac{OX}{AD}=\dfrac{AO}{OS},\quad\text{so}\quad \triangle AOS\sim\triangle ADM.$ Thus $\angle OAS=\angle MAD$ . Combining this with $\angle BAD=\angle CAO$ yields the desired equality. $\blacksquare$

This post has been edited 2 times. Last edited by djmathman, Dec 12, 2015, 10:01 PM

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Kayak

#9 h Jan 4, 2019, 8:35 PM • 1 Y

Y by Adventure10

Let $M_B, M_C$ be the midpoint of $\overline{AC}, \overline{AB}$ respectively. Now $\angle PQO = 90 - \angle QAM_C = \angle ACB$ , similarly $\angle QPO = \angle ABC$ and thus $\Delta ABC \sim \Delta OPQ$ . Let $X$ be the Miquel point of $\Delta M_BM_CD$ w.r.t $\Delta ABC$ , and observe $X$ is the center of spiral similarity $f$ mapping $(A,B,C,O,X) \overset{f}{\mapsto} (O,P,Q,S,X)$ . Since $f(\overline{PQ}) = \overline{BC}$ , $f$ rotates things by $90$ . So $\angle SXO = 90$ , and thus $\angle AXS = \angle AXO + \angle SXO = 90 + 90 = 180$ , and thsu $A,X,S$ are colinear.

Now let $\Phi(Y)$ denote the image of $Y$ after a $\sqrt{bc}$ inversion, then a reflection along $A$ -angle bisector, and then a homothety at $A$ with factor $\frac{1}{2}$ . We have $\Phi(M_B) = C, \Phi(M_C) = B, \Phi(B) = M_C, \Phi(C) = M_B, \Phi(D) = O$ , and thus clearly $\Phi(E) = M$ ('coz $M_C,O,M,B$ are conyclic).

Thus $AX$ and $AM$ are isogonal and we're done.

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#10 h Jan 4, 2019, 8:51 PM • 10 Y

Y by Kayak, AlastorMoody, RudraRockstar, amar_04, khina, 606234, hakN, IAmTheHazard, Adventure10, Mango247

this problem is the proof that russians have time machines. how could they know about 2017G3 in 2010 otherwise??

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#12 h Jun 22, 2021, 6:53 PM

Y by

$\frac{IS}{SJ}=\frac{AI}{JF} \leftrightarrow \frac{OM}{OK}=\frac{KJ}{JF} \leftrightarrow \frac{OK}{OM}=\frac{KF}{KJ}-1 \leftrightarrow \frac{MK\cdot KJ}{OM}=KF$ We have $t=BF$ , $MK=h$ , $AP=\frac{c^2}{2h}$ and $AQ=\frac{b^2}{2h} \implies KJ=AI=\frac{b^2+c^2}{4h}$ , $OM=R\cdot cos(A)$ . Thus it is enough to prove that $\frac{b^2+c^2}{4R\cdot cos(A)}=h+tsin(A) \leftrightarrow t=\frac{a}{2cos(A)}$
$x=\angle BAF$ , $t=FA\frac{sinx}{sinc}=FA\frac{a}{2AM}=\frac{a}{2cos(A)}$
In the last equation we used sin law in $FBA$ and $MAC$ . We also used the fact that $\frac{AM}{AF}=cos(A)$ which can be proved by noticing $AMC\sim AFR$ .

Attachments:

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#13 h Feb 3, 2022, 4:03 PM

Y by

I have solution from sinus. But its easy to chase angels and sinus is very similar to it anyway

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662 posts

#14 h Nov 14, 2023, 5:02 PM • 1 Y

Y by Rounak_iitr

Amazing problem.
$[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -11.028184276513091, xmax = 7.651231623973653, ymin = -9.68127833226805, ymax = 5.291831456807111; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen qqwuqq = rgb(0,0.39215686274509803,0); draw((-0.68,1.85)--(-2.2,-2.65)--(6.9,-3.07)--cycle, linewidth(2) + zzttqq); /* draw figures *//* special point *//* special point *//* special point */ draw((-0.68,1.85)--(-2.2,-2.65), linewidth(0.8)); draw((-2.2,-2.65)--(6.9,-3.07), linewidth(0.8)); draw((6.9,-3.07)--(-0.68,1.85), linewidth(0.8)); draw(circle((2.403617451019996,-1.6982885612334209), 4.700962486344153), linewidth(0.8)); /* special point *//* special point *//* special point *//* special point *//* special point */ draw(circle((-0.1176669679699406,-3.8893946393365), 3.340332463534383), linewidth(0.8) + fuqqzz); /* special point */ draw((-0.68,1.85)--(-0.1176669679699406,-3.8893946393365), linewidth(0.8)); /* special point */ draw(circle((-1.496959200161584,-1.6341160035009759), 1.235429747326925), linewidth(0.8) + qqwuqq); draw(circle((2.9038224601090685,-5.077180030970174), 4.4719354430776574), linewidth(0.8) + qqwuqq); /* special point */ draw((-2.2,-2.65)--(-0.7939184003231672,-0.6182320070019524), linewidth(0.8)); draw((-1.0923550797818622,-7.084360061940349)--(6.9,-3.07), linewidth(0.8)); draw(circle((0.8618087255099981,0.07585571938328961), 2.3504812431720765), linewidth(0.8)); draw((-0.68,1.85)--(2.403617451019996,-1.6982885612334209), linewidth(0.8)); draw((-0.30633208990913463,-1.9638033476805354)--(2.403617451019996,-1.6982885612334209), linewidth(0.8)); draw((2.403617451019996,-1.6982885612334209)--(-0.1176669679699406,-3.8893946393365), linewidth(0.8)); draw((-1.44,-0.4)--(2.403617451019996,-1.6982885612334209), linewidth(0.8)); draw((2.403617451019996,-1.6982885612334209)--(-1.0923550797818622,-7.084360061940349), linewidth(0.8)); draw((2.403617451019996,-1.6982885612334209)--(3.11,-0.61), linewidth(0.8)); draw((-1.0923550797818622,-7.084360061940349)--(-0.68,1.85), linewidth(0.8)); /* dots and labels */ label("$A$", (-0.6,1.93), NE * labelscalefactor); label("$B$", (-2.12,-2.57), NE * labelscalefactor); label("$C$", (6.98,-2.99), NE * labelscalefactor); label("$O$", (2.48,-1.61), NE * labelscalefactor); label("$P$", (-0.72,-0.53), NE * labelscalefactor); label("$Q$", (-1.02,-7.01), NE * labelscalefactor); label("$M$", (-1.36,-0.33), NE * labelscalefactor); label("$N$", (3.18,-0.53), NE * labelscalefactor); label("$S$", (-0.04,-3.81), NE * labelscalefactor); label("$D$", (-0.82,-2.63), NE * labelscalefactor); label("$Q_A$", (-0.22,-1.89), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Let $M$ denote the midpoint of $AB$ , $N$ denote the midpoint of $AC$ , and let $Q_A$ denote the $A$ -dumpty point.
Claim 1: $\triangle OPQ \sim \triangle ABC$
Proof
Since both the triangles are positively similar , therefore there exists a point $Q_A$ such that it takes the former triangle to the latter.
Claim 2: $Q_A$ is the $A$ -dumpty point
Proof
Claim 3: $A-Q_A-S$
Proof
Remark

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Hertz

#15 h Nov 14, 2023, 7:14 PM

Y by

Let $F$ and $G$ be midpoints of $AB$ and $AC$ respectively. Let $K$ be the midpoint $PQ$ .

$\angle POQ = 180 - \angle AOF - \angle AOG = 180 - \angle C - \angle B = \angle A$

Since triangle $AFP$ and triangle $ADB$ are similar we get $\angle QPO = \angle AFP = \angle B$

So we get triangle $OPQ$ and triangle $ABC$ are similar in that orientation.

$\angle SOA = \angle FOA + \angle SOP = \angle C + 90 - \angle C = 90$ So $AO$ is tangent to ( $OPQ$ ), thus $AOST$ is cyclic.

$\angle SAK = \angle SOK = \angle OAM$

$\angle BAS = \angle BAD + \angle SAK = \angle OAC + \angle OAM = \angle CAM$

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53 posts

mcmp

#16 h Nov 8, 2024, 9:25 AM

Y by

Woah Russian Geo :love:

:love:

:love:

I love it when I can just nuke a problem with a technique that appears seven whole years after the problem (in the guise of ISL 2017/G3

).

Rename $M\to M_a$ , let the midpoint of $PQ$ be $M$ , and let the tangent to $(ABC)$ at $A$ intersect $\overline{BC}$ at $A’$ . Let $M_b$ , $M_c$ be the midpoints of $AC$ , $AB$ resp.
Diagram
Imma first show that $AO$ tangent to $(OPQ)$ ; to do this, notice that it would suffice to show that $AP\cdot AQ=AO^2$ . However notice that $M_cPDB$ and $M_bPDC$ are cyclic because of the right angles, so $AP\cdot AD=AM_c\cdot AB$ and $AQ\cdot AD=AM_b\cdot AC$ . Hence:
$\begin{align*} AP\cdot AQ&=\frac{AM_c\cdot AM_b\cdot AB\cdot AC}{AD^2}\\ &=\frac{AB^2\cdot AC^2}{4AD^2}\\ &=\frac{A^2}{AD^2\sin^2(\angle BAC)}\\ &=\frac{BC^2}{\sin^2(\angle BAC)}\\ &=AO^2 \end{align*}$ the last line following from the law of sines. Hence we have the tangency. Now $\measuredangle OPQ=\measuredangle M_cPD=\measuredangle M_cBD=\measuredangle ABC$ , and similarly with $\measuredangle OQP=\measuredangle ACB$ , so $A’ABCM_a\stackrel{+}{\sim}AOPQM$ . The rest is now trivial. $\measuredangle SAD=\measuredangle SAM=\measuredangle OA’M_a=\measuredangle OAM_a$ , so $AM_a$ and $AS$ are conjugate in $\angle DAO$ ; since $AD$ and $AO$ conjugate in $\angle BAC$ , they share the same angle bisector and hence $AM_a$ and $AS$ are conjugate in $\angle BAC$ as desired. $\square$

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