with the radical center

Question

$AD, BE, CF$ are the three heights of $\Delta ABC$, and they concurrent at $H$. $P, Q, R$ are the incenters of $\Delta AQR, \Delta > BRP, \Delta CPQ$, respectively. $\odot (BPR)$ and $ \odot (CPQ) $ intersect with $ \odot (AQR)$ again at $K$ and $L$, respectively. $\odot (BPR)$ and $ \odot (CPQ) $ intersect with $ \odot (ABC)$ again at $S$ and $T$, respectively. $\odot(SKH)$ and $ \odot (LTH)$ intersect with each other again at $X$ Show that:

1) $K,S,T,L$ are cyclic

2) $P,H,X$ are colinear

What I thought: $P$ is the incenter of $\Delta AEF$, where $BE \perp AC$ at $E$, $CF \perp AB$ at $ F$. $Q$ and $R$ are defined analogously.

Answer 1

It is clear that the "composer" of the problem constructed a very special situation for (1), and did not want to tell us which are the "minimal requirements" for the points $A,B,C;P,Q,R$, so that the result in (1) is still holding. The solution of the problem goes then through the step of finding such a "minimal configuration" of points.

(Personally i do not consider is it a good idea to make out a clean geometry situation an overloaded situation by passing to very particular cases, so that it becomes a puzzle.)

So in order to solve the puzzle, i have to extract the following situation, with points having names obtained by applying an inversion $W\to W^*$ centered in $A$ starting from the given situation.

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Result 1.1 Let $B^*,C^*,R^*,Q^*$ be four points on a circle in the plane. Consider $P^*$, a further point. We construct the four intersections:

• $K^*$ is the second intersection of line $Q^*R^*$ with circle $(B^*P^*R^*)$.
• $L^*$ is the second intersection of line $Q^*R^*$ with circle $(C^*P^*Q^*)$.
• $S^*$ is the second intersection of line $B^*C^*$ with circle $(B^*P^*R^*)$.
• $T^*$ is the second intersection of line $B^*C^*$ with circle $(C^*P^*Q^*)$.

Then these four intersection points lie on a circle.

Proof: Let $Z$ be the intersection of $P^*Q^*$ and $B^*C^*$. We will denote by the small capital letter $x$ the lenghth of $ZX^*$, so for instance $b=ZB^*$. The using the powers of $Z$ w.r.t. the given circles we have $bs=kr$, $ct=ql$, and $bc=qr$. From the first two relations we get $$ bcst=qrkl\ ,$$ and dividing with the third one, $st=kl$, which means that the four constructed intersections are on a circle.

$\square$

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Result 1.2 Let $ABC$ be a triangle. Let $D,E,F$ be the projections of the vertices $A,B,C$ on the opposite sides, let $H$ be the orthocenter $AD\cap BE\cap CF$. We denote by $P,Q,R$ the incenters of the triangles $\Delta AEF$, $\Delta BFD$, $\Delta CDE$. In particular, the intersection of angle bisectors $AP\cap BQ\cap CR$ is $I$, the incenter of $\Delta ABC$.

Let the angles $\hat A$, $\hat B$, $\hat C$ in $\Delta ABC$ have the measures $2x,2y,2z$.

Then we have $$IA\cdot IP=IB\cdot IQ=IC\cdot IR$$ and the angles in $\Delta PQR$, further delimited by $PI, QI, RI$ are as in the following picture:

Bonus: $I$ is the orthocenter of $\Delta PQR$. (Since for instance $\angle IQR=z$, and $\angle QRP =x+y$, and $x+y+z=90^\circ$, so $I$ is on the height drawn from $Q$.)

Proof: The marked angles based in $A,B,C;D,E,F$ having measures $x,y,z$ are clear.

We show now the metric relation. First of all, the triangles $\Delta ABC$ and $\Delta AEF$ are similar (in this order of vertices), so the constructed distances from $A$ to the incenters are also respecting the same similarity factor, $$ \frac{AP}{AI}= \frac{AE}{AB}=\cos A=\cos (2x)\ . $$ This implies $$ IA\cdot IP=(1-\cos (2x))\cdot IA^2=2\sin ^2 x\cdot IA^2\ . $$ Similar relations for the other two products. The claimed relation follows now from the sine theorem applied for (two of) the triangles $\Delta IBC$, $\Delta ICA$, $\Delta IAB$. For instance, in the first triangle we have $$ \frac{IB}{\sin z}=\frac{IC}{\sin y}\ . $$ From the obtained metric relation, the following quadrilaterals are cyclic: $BCRQ$, $CAPR$, $ABQP$. This explains the distribution of the marked angles inside $\Delta PQR$.

$\square$

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Proof of (1): We have shown above in the Result 1.2 that $B,C,Q,R$ are on a circle. Let us apply an inversion centered in $A$, denoted by $*$. Then the image points $B^*,C^*,Q^*,R^*$ are on a circle and we are in the position to apply the Result 1.1, showing that $K^*,S^*, T^*, L^*$ are on a circle. These points, constructed as in Result 1.1 are the images of the points $K,S,T,L$ from the OP, since by inversion, we have the following transformations:

• $A$ goes to an "infinity point" $A^*=\infty$.
• The circle $(AQR)$ becomes the (projective) line $\infty Q^*R^*$.
• The circle $(ABC)$ becomes the (projective) line $\infty B^*C^*$.
• The circle $(BPR)$ becomes the circle $ B^*P^*R^*$.
• The circle $(CPQ)$ becomes the circle $ C^*P^*Q^*$.

And intersections correspond to intersections.

We apply the same inversion "back" to see that the circle $(K^*S^*T^* L^*)$ transforms in a circle $KSTL$.

$\square$

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Now for the second part we also have to isolate a simple configuration first, this will be Result 2.1, then show that the hypotesis of this result is matched in the complicated figure we start with.

Result 2.1: Let $S,K;T,L$ be four points on a circle or on a line. Let $C,\Gamma$ be two (different) circles through $S,K$. Let $C',\Gamma'$ be two (different) circles through $T,L$. Let $P,Y$ be the intersection points of $C,C'$. Let $H,X$ be the intersection points of $\Gamma,\Gamma'$.

Then the four constructed points $P,Y,H,X$ are on a circle or on a line.

Proof: After an inversion, we can and do assume w.l.o.g. that the four starting points $S,K,T,L$ are on a line. We now apply a further inversion with center in $S$. It moves $S$ to an infinity point and the two circles $C,\Gamma$ through $S$ are mapped into lines. We assume w.l.o.g. that we have this situation, and do not introduce notations to show the difference. The picture is now as follows:

From the transfer of powers w.r.t. the two remained circles and the point $K$, using the intermediate power $KT\cdot KL$, $$ KP\cdot KY = KT\cdot KL =KH\cdot KX $$ we obtain immediately $P,Y,H,X$ on a circle.

$\square$

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It is now clear, that we "only" have to show for (2) in the OP the following...

Result 2.2: Reconsider the situation from Result 1.2. The point $P$ is one of the intersection points of the circles $(BKPRS)$ and $(CLPQR)$. Let $Y$ be the other intersection.

Then the points $P,Y,H$ are colinear.

Proof: Here i am missing a synthetic idea, so in order to buy my freedom and submit a solution the proof will be an analytic one, using barycentric coordinates. Computations are straightforward.

Let $a,b,c$ be the (lengths of the) sides of $\Delta ABC$.

A point $W$ has (true) barycentric coordinates $x,y,z$ if $x+y+z=1$ and $W=xA+yB+zC$ (afix computation), i.e. iff for one or any reference point $O$ in the plane we have (vectorially) $OW=xOA+yOB+zOC$. Sometimes, rather often, we use the notation $(x:y:z)$ for the coordinates, in this case one has to divide by the sum $x+y+z\ne 0$, so $(x:y:z)$ is $\displaystyle \left( \frac x{x+y+z}, \frac y{x+y+z}, \frac z{x+y+z} \right) $.

Barycentric coordinates are collected for instance in the ETC (encyclopedia of triangle centers). Central triangles are also important in the context. (The triangle $PQR$ is one of them, i can not identify which on right now.)

The incenter $I=X(1)$ has (inhomogenous) barycentric coordinates $(a:b:c)$. The point $A$ has coordinates $(1:0:0)$. The point $P$ is on $IA$ and we have already computed $AP:AI=\cos A$, so $$ \begin{aligned} P &=\cos A\cdot I + (1-\cos A)\cdot A\\ &=\cos A\left(\ \frac a{a+b+c},\ \frac b{a+b+c},\ \frac c{a+b+c}\ \right) + (1-\cos A)(1,0,0) \\ &=\left(\ 1- \cos A\frac {b+c}{a+b+c},\ \cos A\frac b{a+b+c},\ \cos A\frac c{a+b+c}\ \right) \\ &=\left(\ \frac {a+b+c}{\cos A}-(b+c)\ :\ b\ :\ c\ \right) \\ &=(x_P:b:c)\ . \end{aligned} $$ The expression $x_P$ is defined by the last line.

Similar computations give $Q=(a:y_Q:c)$, and $R=(a:b:z_R)$ with similar expressions for $y_Q$, $z_R$.

(We can replace at any point $\cos A$ by the rational expression in $a,b,c$ extracted from $a^2=b^2+c^2-2bc\cos A$. The other cosine values for $B,C$, too.)

The general homogenous equation of a circle in barycentric coordinates $(x:y:z)$ is $$ 0=-a^2yz-b^2xz-c^2xy +(ux+vy+wz)(x+y+z)\ . $$ We need the specific equation for the circle $(CPQ)$ first, and denote by $u_1,v_1,w_1$ the matching coefficients. (I.e. the specific values for the unknowns $u,v,w$ that verify.) Since $C(0,0,1)$ verifies, we get for instance immediately $w_1=0$. The other values are not so simple, so i decided to get computer help, here sage. For the circle $(BPQ)$ we obtain coefficients $u_2,v_2,w_2$, as a matter of notation, and this time $v_2=0$ since $B(0,1,0)$ verifies the equation.

The following sage code checks in a simple manner, that the orthocenter $$ H=(\tan A:\tan B:\tan C) = \left( \frac a{\cos A}: \frac b{\cos B}: \frac c{\cos C} \right) $$ is on the line $(u_1-u_2)x+(v_1-v_2)y+(w_1-w_2)z=0$, which is the intersection of the two circles, obtained by formally subtracting there equations. (And dividing by $x+y+z\ne 0$.)

var('a,b,c,u,v,w');

def eq(point):
    x, y, z = point
    return -a^2*y*z -b^2*z*x -c^2*x*y + (u*x + v*y + w*z)*(x+y+z)

cosA = (b^2 + c^2 - a^2) / (2*b*c)
cosB = (c^2 + a^2 - b^2) / (2*c*a)
cosC = (a^2 + b^2 - c^2) / (2*a*b)

xP = (a+b+c)/cosA - b - c
yQ = (a+b+c)/cosB - c - a
zR = (a+b+c)/cosC - a - b

P = (xP, b, c)
Q = (a, yQ, c)
R = (a, b, zR)

A = (1,0,0)
B = (0,1,0)
C = (0,0,1)

solCPQ = solve( [eq(P), eq(Q), eq(C)], [u,v,w], solution_dict=1 )[0]
solBPR = solve( [eq(P), eq(R), eq(B)], [u,v,w], solution_dict=1 )[0]

u1, v1, w1 = solCPQ[u], solCPQ[v], solCPQ[w]
u2, v2, w2 = solBPR[u], solBPR[v], solBPR[w]

E = (u1-u2)*a/cosA + (v1-v2)*b/cosB + (w1-w2)*c/cosC
E.simplify_full()

The above delivers a clean zero. This was the check.

$\square$

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I am not proud of this solution, but somehow i have to part in peace with the whole ticket. The above is in my eyes a valid solution. It may be desirable in such situations to have a less analytic solution, best would be to provide a synthetic solution, so that esthetic criteria are not hurt, well, if i find a short one i will insert it next days.

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Proof of (2):

We can now conclude using the two results above (if this is not already done). We apply Result 2.1 mot-a-mot for the letters from the OP, so the four points $P,Y;H,X$ are either on a line or on a circle. From Result 2.2 the points $P,H,Y$ are on a line, so the circle is a line. (And we have four points known on it.)

$\square$

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A final comment. After the whole procession we have a result which is a dead end, so the whole effort is not really a decent reward. For (2), the points $S,T,K,L$ are so complicated that i cannot see a possibility to show (2) without introducing $Y$. The problem could do this for us and claim more, not doing so is part of a puzzle.