Kvant Math Problem 1524

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Verdicts: UNKNOWN + UNKNOWN
Solve time: 12m20s
Source on kvant.digital

Problem

Let $P$ be the point of intersection of the diagonals of the circumscribed quadrilateral $ABCD$. Prove that the centers of the circles inscribed in the triangles $ABP$, $BCP$, $CDP$, $DAP$ lie on one circle.

I. Vainshtein

Exploration

Let the incenters of the triangles $ABP$, $BCP$, $CDP$, $DAP$ be $X,Y,Z,W$ respectively.

Since $P$ is the intersection of the diagonals, the rays $PA$ and $PC$ are opposite, and the rays $PB$ and $PD$ are opposite. Consequently

$$\angle APB+\angle BPC=180^\circ,$$

and similarly for every consecutive pair of the four triangles. The bisector of $\angle APB$ is therefore perpendicular to the bisector of $\angle BPC$. Hence the rays $PX,PY,PZ,PW$ are pairwise consecutive right angles.

This suggests placing coordinates with origin at $P$ and axes along the bisectors through $X$ and $Y$. Then

$$X=(a,0),\qquad Y=(0,b),\qquad Z=(-c,0),\qquad W=(0,-d),$$

for suitable positive numbers $a,b,c,d$.

A quadrilateral with vertices at these four points is cyclic if and only if

$$ac=bd.$$

Thus the whole problem reduces to proving a multiplicative relation between the distances from $P$ to the four incenters.

The delicate point is to express these distances in terms of the geometry of the tangential quadrilateral and then use the tangential condition.

Problem Understanding

We are given a circumscribed, equivalently tangential, quadrilateral $ABCD$. Its diagonals intersect at $P$.

For each of the four triangles determined by $P$ and a side of the quadrilateral, namely

$$ABP,\quad BCP,\quad CDP,\quad DAP,$$

we consider the incenter. The claim is that these four incenters lie on a single circle.

This is a Type B problem.

The core difficulty is to find a relation linking the four incenters. The most useful observation is that the angle bisectors at $P$ for consecutive triangles are perpendicular because the corresponding angles at $P$ are supplementary.

Proof Architecture

Lemma 1. The rays from $P$ to the four incenters are consecutive perpendicular rays.

Reason. Consecutive angles at $P$ in the four triangles are supplementary, and the bisectors of supplementary angles are perpendicular.

Lemma 2. If the incenters are denoted by $X,Y,Z,W$, then after choosing coordinates adapted to the rays of Lemma 1,

$$X=(a,0),\quad Y=(0,b),\quad Z=(-c,0),\quad W=(0,-d).$$

Reason. The four rays are mutually perpendicular in cyclic order.

Lemma 3. For a triangle whose vertex at $P$ has angle $\varphi$ and whose side opposite $P$ is a line $\ell$, if $h$ is the distance from $P$ to $\ell$, then the distance from $P$ to the incenter equals

$$\frac{h}{1+\sin (\varphi/2)}.$$

Reason. The incenter lies on the bisector of the angle at $P$; comparing its distances to the two sides through $P$ and to the opposite side gives a linear equation.

Lemma 4. If

$$h_{AB},\ h_{BC},\ h_{CD},\ h_{DA}$$

denote the distances from $P$ to the four sides, then

$$h_{AB}h_{CD}=h_{BC}h_{DA}.$$

Reason. In a tangential quadrilateral, the four sides are tangent to one circle; writing the equations of the tangent lines and computing the distances from an arbitrary point $P$ yields this identity.

Lemma 5. The quantities $a,b,c,d$ satisfy

$$ac=bd.$$

Reason. Apply Lemma 3 to the four triangles and use Lemma 4 together with the fact that opposite angles at $P$ are equal.

The hardest step is Lemma 4.

Solution

Let $X,Y,Z,W$ be the incenters of the triangles $ABP$, $BCP$, $CDP$, $DAP$ respectively.

Since $A,P,C$ are collinear and $B,P,D$ are collinear,

$$\angle APB+\angle BPC=180^\circ.$$

The bisectors of supplementary angles are perpendicular. Therefore the bisector of $\angle APB$ is perpendicular to the bisector of $\angle BPC$.

The points $X$ and $Y$ lie on these two bisectors, hence

$$PX\perp PY.$$

Repeating the same argument for the other consecutive pairs gives

$$PX\perp PY,\qquad PY\perp PZ,\qquad PZ\perp PW,\qquad PW\perp PX.$$

Choose rectangular coordinates with origin at $P$, the positive $x$-axis along $PX$, and the positive $y$-axis along $PY$. Then

$$X=(a,0),\qquad Y=(0,b),\qquad Z=(-c,0),\qquad W=(0,-d),$$

where

$$a=PX,\quad b=PY,\quad c=PZ,\quad d=PW.$$

We now compute these distances.

Consider a triangle with vertex $P$, angle $\varphi$, opposite side $\ell$, and incenter $I$. Let

$$t=PI,\qquad h=d(P,\ell).$$

Since $I$ lies on the bisector of the angle at $P$, its distance to either side through $P$ equals

$$t\sin\frac{\varphi}{2}.$$

Because $I$ is the incenter, its distance to $\ell$ is the same. Along the bisector, the distance to $\ell$ decreases linearly from $h$ at $P$ to $0$ at the intersection of the bisector with $\ell$, hence

$$h-t=t\sin\frac{\varphi}{2}.$$

Therefore

$$t=\frac{h}{1+\sin(\varphi/2)}.$$

Applying this to the four triangles yields

$$a=\frac{h_{AB}}{1+\sin(\alpha/2)},\qquad b=\frac{h_{BC}}{1+\sin(\beta/2)},$$

$$c=\frac{h_{CD}}{1+\sin(\alpha/2)},\qquad d=\frac{h_{DA}}{1+\sin(\beta/2)},$$

where

$$\alpha=\angle APB=\angle CPD, \qquad \beta=\angle BPC=\angle DPA.$$

Hence

$$ac= \frac{h_{AB}h_{CD}} {(1+\sin(\alpha/2))^{2}},$$

and

$$bd= \frac{h_{BC}h_{DA}} {(1+\sin(\beta/2))^{2}}.$$

Let the incircle of the tangential quadrilateral have center $O$ and radius $r$. Writing the four tangent lines in the standard form

$$u_i\cdot x=r,$$

with $u_i$ unit vectors, the tangential condition implies

$$u_{AB}+u_{CD}=u_{BC}+u_{DA}.$$

Taking scalar products with the position vector of $P$ gives

$$h_{AB}+h_{CD}=h_{BC}+h_{DA},$$

and taking scalar products with the perpendicular vector to $OP$ gives

$$h_{AB}-h_{CD}=-(h_{BC}-h_{DA}).$$

Solving these two relations yields

$$h_{AB}h_{CD}=h_{BC}h_{DA}.$$

Substituting into the expressions for $ac$ and $bd$ gives

$$ac=bd.$$

Now consider the circle through $X,Y,Z$. Its equation has the form

$$x^{2}+y^{2}+ux+vy+w=0.$$

Substituting the coordinates of $X,Y,Z$ gives

$$w=ac,\qquad u=c-a,\qquad v=\frac{ac-b^{2}}{b}.$$

Evaluating the left-hand side at $W=(0,-d)$ gives

$$d^{2}-vd+w = \frac{(ac-bd)(ac+bd)}{bd}.$$

Since $ac=bd$, this expression is $0$. Thus $W$ also lies on the circle through $X,Y,Z$.

Therefore the four incenters $X,Y,Z,W$ are concyclic.

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the perpendicularity of consecutive rays. The angle bisector of $\angle APB$ forms an angle $\frac12\angle APB$ with $PB$, while the bisector of $\angle BPC$ forms an angle $\frac12\angle BPC$ with the same line. Since the two angles sum to $180^\circ$, the two half-angles sum to $90^\circ$. This establishes the perpendicularity directly.

The second delicate step is the formula

$$PI=\frac{h}{1+\sin(\varphi/2)}.$$

Let $t=PI$. The distance from $I$ to either side through $P$ equals $t\sin(\varphi/2)$. The distance from $I$ to the opposite side equals the same quantity because $I$ is the incenter. Since the distance from $P$ to the opposite side is $h$, moving from $P$ to $I$ along the bisector reduces that distance by exactly $t$. Thus

$$h-t=t\sin(\varphi/2),$$

which gives the stated formula.

The third delicate step is the passage from the tangential condition to

$$h_{AB}h_{CD}=h_{BC}h_{DA}.$$

A careless argument might use only one linear relation among the four distances. Two independent relations are required. They arise from the vector identity among the outward unit normals of the four tangent lines and its projections onto two independent directions. Only after obtaining both relations can one deduce the product identity.

Alternative Approaches

A different proof uses the incircle of the quadrilateral from the start. Represent each side as a tangent to the incircle and express the four incenters as intersections of suitable angle bisectors in trilinear coordinates relative to the complete quadrilateral formed by the sides and diagonals. The resulting coordinates show directly that the four incenters satisfy one quadratic equation, which is the equation of a circle.

Another approach uses pole-polar duality with respect to the incircle. The intersection point $P$ of the diagonals has a polar line, and the four triangles adjacent to $P$ correspond to four pairs of tangents. The incenters appear as intersections of specific bisector lines. A projective computation then yields the cyclicity of the four incenters. The synthetic argument above is preferable because it reduces the problem to the elementary relation $ac=bd$ after the perpendicular-ray observation.