Let $M$ be the midpoint of $BC$ of triangle $ABC$. The circle with diameter $BC$, $\omega$, meets $AB,AC$ at $D,E$ respectively. $P$ lies inside $\triangle ABC$ such that $\angle PBA=\angle PAC, \angle PCA=\angle PAB$, and $2PM\cdot DE=BC^2$. Point $X$ lies outside $\omega$ such that $XM\parallel AP$, and $\frac{XB}{XC}=\frac{AB}{AC}$. Prove that $\angle BXC +\angle BAC=90^{\circ}$.
Problem
Source: China TST Test 2 Day 2 Q5
Tags: geometry, China TST
anantmudgal09
11.03.2019 23:07
I am probably missing something, but the condition $2PM \cdot DE=BC^2$ is redundant, since the previous two angle conditions force $P$ to be the midpoint of $A$-symmedian chord, right?
rmtf1111
11.03.2019 23:14
anantmudgal09 wrote: I am probably missing something, but the condition $2PM \cdot DE=BC^2$ is redundant, since the previous two angle conditions force $P$ to be the midpoint of $A$-symmedian chord, right? not in every triangle this condition is satisfied
K_Mirrorheart
12.03.2019 09:48
Well this problem has got a nice geometric solution, and I would point out that $A, D, E$ are largely irrevelant.
61plus
15.03.2019 08:58
Claim 1: $P$ is the midpoint of the $A$-symmedian chord. This is true for $P$ satisfying the condition $P$ lies inside $\triangle ABC$ such that $\angle PBA=\angle PAC, \angle PCA=\angle PAB$:
Clearly the point $P$ is unique, as it is the second intersection of the circle passing through $A,B$ tangent at $A$, and the circle passing through $A,C$ tangent at $A$.
Let $S$ be the intersection of the tangents at $B,C$; $P'$ the midpoint of the $A$-symmedian chord. $O$ is the circumcenter of $\triangle ABC$. Then $OP'\perp AS$ so $OP'BSC$ concyclic. Then $\angle BP'S=\angle A$, so $\angle PB'A=\angle A -\angle P'AB=\angle P'AC$ and similarly $\angle P'CA=\angle P'AB$, thus $P'\equiv P$.
Claim 2: $MP=BS$.
Note that $2DE = 4\cos{A}\cdot MD=4\cos{A}\cdot MB$. Thus $PM=\frac{BM}{\cos{A}}=BS$.
The condition $\frac{XB}{XC}=\frac{AB}{AC}$ implies that $X$ lies on the $A$-Apollonius Circle of $ABC$. However the line passing through $M$ parallel to $AP$ should intersect this circle at two points, so we call the other intersection point $Y$ which satisfy the same properties as $X$, except that it may be in $\omega$. Claim 3: $\angle BXC+\angle BYC=180^{\circ}$.
Reflect $Y$ across $M$ to $Y'$. So we want $BY'CX$ concyclic which means that $BM\cdot MC=XM\cdot MY' = MX\cdot MY$. But this is true as the power of $M$ to the $A$-Apollonius Circle is $MB^2$.
So we want to show that $\angle BYC=90+\angle A$, which is equivalent to $Y$ lying on the circle with radius $SB$, or $SY=SB$. Instead now, we construct the point $Y'$ such that $SY'=SB$, $MY'//AP$ and $Y',X$ are on the same side of $BC$. Note that $SY'=MP$, so $MY'PS$ is an isosceles trapezoid thus $O,Y',P$ must be collinear. Now if we show that $Y'$ lies on the $A$-Apollonius Circle then we are done. Let the intersection of $AS,BC$ be $L$. Then it is well known that $\frac{BL}{CL}=\frac{AB^2}{AC^2}$. So we want $Y'L$ to be the symmedian of $\triangle Y'BC$ which will give $\frac{Y'B}{Y'C}=\sqrt{\frac{BL}{CL}}=\frac{AB}{AC}$. Note that $OB,OC$ are tangent to the circumcircle of $\triangle Y'BC$, so our desired result is equivalent to showting $O,Y',L$ are collinear. This is true because $SY'^2=SB^2=SL\cdot SP=SM\cdot SO$ and thus $\angle SY'L=\angle SPY'=180^{\circ}-\angle SMY'=180^{\circ}-\angle SY'O$. So $Y\equiv Y'$ and we are almost done. Finally we want to show that $Y$ cannot be the point $X$, as it lies inside $\omega$. However this is clear as $\angle BY'C=90^{\circ}+\angle A>\angle BDC$.
Pathological
16.03.2019 05:48
A somewhat different finish, after getting Claims 1-3 of the above solution: Let $T = BB \cap CC$, $\Gamma$ be the $A-$Apollonius circle of $\triangle ABC$, and $X' = MX \cap \Gamma.$ Observe that $X \in \Gamma.$ It is well-known that $P$ is the midpoint of the $A-$symmedian chord in $\triangle ABC.$ Now, observe that $2*PM * DE = BC^2 \Leftrightarrow PM * 2a \cos{\angle A} = a^2 \Leftrightarrow PM = BT.$ Letting $Y, Z$ be the feet of the internal, external angle-bisector of $\angle BAC$ onto $BC$, respectively, we know that $(B, C; Y, Z) = -1 \Rightarrow MY * MZ = MB^2$ by a well-known property of harmonic ranges. Now, this therefore implies that $MX * MX' = MB^2$ by Power of the Point, and so hence $\triangle MX'B \sim \triangle MBX, \triangle MX'C \sim \triangle MCX.$ These then imply that $\angle BXC + \angle BX'C = 180,$ and so hence $\angle BXC < 90$ clearly implies that $BX > BX'.$ Therefore, we need now to show that $\angle BX'C = 90 + \angle A,$ which is equivalent to $TX' = TB$. In order to show that, we will show that $PX'MT$ is an isosceles trapezoid, which would then imply that $TX' = PM = TB,$ as desired. To show this, we only need to show that $PX' = MT.$ Notice that $PX'^2 - PM^2 = ZX'^2 - ZM^2$ by Carnot, where $Z$ is the foot of the altitude from $P$ to $XX'.$ However, since $P$ is the midpoint of the $A-$symmedian chord and $XX' || AP,$ we know that $ZX'^2 - ZM^2 = (ZX' - ZM)(ZX'+ZM) = (-X'M)(XM) = -MB^2.$ Therefore, we have that $PX'^2 - PM^2 = -BM^2,$ which hence implies that $PX'^2 = PM^2 - BM^2 = BT^2 - BM^2 = MT^2,$ hence implying the problem. $\square$
Psimo
06.04.2019 12:11
61plus wrote: Claim 1: $P$ is the midpoint of the $A$-symmedian chord. This is true for $P$ satisfying the condition $P$ lies inside $\triangle ABC$ such that $\angle PBA=\angle PAC, \angle PCA=\angle PAB$:
Clearly the point $P$ is unique, as it is the second intersection of the circle passing through $A,B$ tangent at $A$, and the circle passing through $A,C$ tangent at $A$.
Let $S$ be the intersection of the tangents at $B,C$; $P'$ the midpoint of the $A$-symmedian chord. $O$ is the circumcenter of $\triangle ABC$. Then $OP'\perp AS$ so $OP'BSC$ concyclic. Then $\angle BP'S=\angle A$, so $\angle PB'A=\angle A -\angle P'AB=\angle P'AC$ and similarly $\angle P'CA=\angle P'AB$, thus $P'\equiv P$.
Claim 2: $MP=BS$.
Note that $2DE = 4\cos{A}\cdot MD=4\cos{A}\cdot MB$. Thus $PM=\frac{BM}{\cos{A}}=BS$.
The condition $\frac{XB}{XC}=\frac{AB}{AC}$ implies that $X$ lies on the $A$-Apollonius Circle of $ABC$. However the line passing through $M$ parallel to $AP$ should intersect this circle at two points, so we call the other intersection point $Y$ which satisfy the same properties as $X$, except that it may be in $\omega$. Claim 3: $\angle BXC+\angle BYC=180^{\circ}$.
Reflect $Y$ across $M$ to $Y'$. So we want $BY'CX$ concyclic which means that $BM\cdot MC=XM\cdot MY' = MX\cdot MY$. But this is true as the power of $M$ to the $A$-Apollonius Circle is $MB^2$.
So we want to show that $\angle BYC=90+\angle A$, which is equivalent to $Y$ lying on the circle with radius $SB$, or $SY=SB$. Instead now, we construct the point $Y'$ such that $SY'=SB$, $MY'//AP$ and $Y',X$ are on the same side of $BC$. Note that $SY'=MP$, so $MY'PS$ is an isosceles trapezoid thus $O,Y',P$ must be collinear. Now if we show that $Y'$ lies on the $A$-Apollonius Circle then we are done. Let the intersection of $AS,BC$ be $L$. Then it is well known that $\frac{BL}{CL}=\frac{AB^2}{AC^2}$. So we want $Y'L$ to be the symmedian of $\triangle Y'BC$ which will give $\frac{Y'B}{Y'C}=\sqrt{\frac{BL}{CL}}=\frac{AB}{AC}$. Note that $OB,OC$ are tangent to the circumcircle of $\triangle Y'BC$, so our desired result is equivalent to showting $O,Y',L$ are collinear. This is true because $SY'^2=SB^2=SL\cdot SP=SM\cdot SO$ and thus $\angle SY'L=\angle SPY'=180^{\circ}-\angle SMY'=180^{\circ}-\angle SY'O$. So $Y\equiv Y'$ and we are almost done. Finally we want to show that $Y$ cannot be the point $X$, as it lies inside $\omega$. However this is clear as $\angle BY'C=90^{\circ}+\angle A>\angle BDC$. $P$ is the $A$ dumpty point (i'm not pretty sure it's the midpoint of the symmedian, in fact it lies on it ).
bern-1-16-4-13
17.05.2019 19:29
Let $N$ be the intersection point of the tangents to $\left(ABC\right)$ in $B$ and $C$ respectively. Define also $D=AN\cap\left(ABC\right)$.
$2PM\cdot DE=BC^2\Leftrightarrow PM=BN$.
Infact by $AED\sim ABC$ we have $DE=\frac{BC\cdot AE}{AB}=BC\cos\alpha\Rightarrow 2PM=\frac{BC}{2\cos\alpha}=\frac{BM}{\cos\alpha}=BN$ since $\angle{NBM}=\alpha$.
$P$ is the midpoint of $AD$.
To show this let $\Phi$ be an inversion in $A$ with radius $\sqrt{AB\cdot AC}$ followed by a symmetry wrt the internal bisector of $\angle{ABC}$. It's well known that $\Phi\left(D\right)=M$, so it now suffices to show that $\Phi\left(P\right)=A'$ defined as the symmetric of $A$ wrt $M$. It's easy to see that $P$ is defined in the text as intersection between $\Omega$ and $\Gamma$ which are respectively the circle through $A,B$ tangent to $AC$ and the circle through $A,C$ tangent to $AB$. Hence $\Phi\left(P\right)=\Phi\left(\Omega\right)\cap\Phi\left(\Gamma\right)=A'$ since the images of $\Omega$ and $\Gamma$ are the parallel to $AB$ through $C$ and the parallel to $AC$ through $B$.
Now let $O$ be the center of the $A$-Apollonius circle, $Y=XM\cap\left(PMN\right)$ and $\pi$ be the circle through $B$ and $C$ centered at $N$.
$O\in\left(PNMY\right)$ and $Y\in\pi$.
For the first one just note $OP\perp AD$ (by Step 2) and $OM\perp MN$. For the second note that $BN\stackrel{\text{Step 1}}{=}MP=NY$ since $NPYM$ is an isosceles trapezoid (because it is cyclic).
$Y$ lies on the $A$-Apollonius circle (together with $X$ and $D$ of course).
To show this it suffices that $OY$ is tangent to $\pi$, which is clear by Step 3 because we have $\angle{OYN}=\angle{APN}=90$. In particular $AXYD$ is a cyclic trapezoid, hence it is isosceles.
Call now $X'$ the symmetric of $X$ wrt $M$.
$X'\in\pi$.
This is true iff $BN=X'N\Leftrightarrow X'N=MP\stackrel{NP\parallel MX'}{\Leftrightarrow} NPMX'$ is a parallelogram $\Leftrightarrow NP=MX'\Leftrightarrow NP=MX\Leftrightarrow NPXM$ is a parallelogram $\Leftrightarrow PX=NM$ and, remembering that the trapezoid $NPYM$ is isosceles, this last equality is true iff $PX=PY$. But this is a direct consequence of Step 2 and Step 4 which tell us that $P$ is midpoint of $AD$ and that $AXYD$ is isosceles.
Now we are ready to show that $\angle{BXC}=90-\angle{BAC}$. We know that $\angle{BXC}=\angle{CX'B}\stackrel{\text{Step 5}}{=}\frac{\angle{BNC}}{2}=\angle{BNM}=90-\angle{BNC}=90-\angle{BAC}$. $\blacksquare$
nguyenhaan2209
15.07.2019 19:10
Notice that P is the midpoint of symmedian AF. Let (I)=A-Appolonius, if MX tangent to (I) then X on (w) absurd so take Y=2nd intersection. By Maclaurin, BYC+BXC=180. Notice PM.2acosA=a^2 then PM=BS, we want BYC=90+A so retake Y in MX that SY=SB=PM so MS=PY=PX so PX perp BC. Notice YOP=XOP=PXI=IMP=NOP (N=OS-BC) hence Y,O,N collinear (OMNP cyclic and IO.IP=IX^2) so YN is symmedian so YB/YC=(NB/NC)^2=AB/AC so Y also lies on A-appolonius so Y satisfy the condition hence conclusion.