Kvant Math Problem 568

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Problem

The diagonals of a convex quadrilateral $ABCD$ intersect at the point $O$. Prove that

1. if the radii of the circles inscribed in the triangles $AOB$, $BOC$, $COD$, and $DOA$ are equal, then $ABCD$ is a rhombus;
2. if the radii of the circles inscribed in the triangles $ABC$, $BCD$, $CDA$, and $DAB$ are equal, then $ABCD$ is a rectangle.

A. A. Egorov

13th All-Union School Olympiad, Grade 10

Exploration

Let

$$x=AO,\quad y=BO,\quad z=CO,\quad t=DO.$$

For Part 1 the triangles are exactly the four triangles around the intersection of the diagonals. Since the diagonals of a convex quadrilateral intersect, the angles at $O$ in triangles $AOB$ and $COD$ are equal, and the angles at $O$ in triangles $BOC$ and $DOA$ are equal.

For a triangle with sides $a,b$ enclosing angle $\theta$, the inradius is

$$r=\frac{\Delta}{s} =\frac{ab\sin\theta}{a+b+\sqrt{a^{2}+b^{2}-2ab\cos\theta}}.$$

The denominator can be simplified:

$$(a+b+c)(a+b-c) =(a+b)^2-c^2 =2ab(1+\cos\theta),$$

hence

$$a+b+c=\frac{2ab(1+\cos\theta)}{a+b-c}.$$

Using

$$(a+b-c)^2=2ab(1-\cos\theta),$$

we obtain

$$r=(a+b-c)\frac{\sin\theta}{2(1+\cos\theta)} =(a+b-c)\frac{\tan(\theta/2)}2.$$

Thus for fixed $\theta$,

$$r=\frac{\tan(\theta/2)}2,(a+b-c).$$

Now $AOB$ and $COD$ have the same angle $\theta$ at $O$. Equality of their inradii yields

$$x+y-\sqrt{x^2+y^2-2xy\cos\theta} = z+t-\sqrt{z^2+t^2-2zt\cos\theta}.$$

A useful identity is

$$a+b-c=\frac{2ab(1+\cos\theta)}{a+b+c}.$$

Hence equality of the above expressions becomes

$$\frac{xy}{x+y+c_1} = \frac{zt}{z+t+c_2},$$

where $c_1,c_2$ are the corresponding opposite sides.

This still looks messy.

A better route is to introduce

$$u=a+b-c.$$

Since

$$u^2=2ab(1+\cos\theta)-2c(a+b-c),$$

that is not immediately helpful.

Try expressing $u$ via half-angle. Since

$$(a+b-c)^2 =2ab(1+\cos\theta) =4ab\cos^2\frac\theta2,$$

we get

$$a+b-c=2\sqrt{ab}\cos\frac\theta2.$$

This is the crucial simplification. Indeed,

$$(a+b-c)^2 =(a+b)^2-c^2 =2ab(1+\cos\theta).$$

Therefore

$$r=\sqrt{ab}\sin\frac\theta2.$$

This formula is elegant.

Hence for Part 1,

$$r_{AOB}=\sqrt{xy}\sin\frac\alpha2,\qquad r_{COD}=\sqrt{zt}\sin\frac\alpha2,$$

where $\alpha=\angle AOB$.

Equality gives

$$xy=zt.$$

Similarly, from $r_{BOC}=r_{DOA}$,

$$yz=xt.$$

Since all quantities are positive,

$$\frac{xy}{yz}=\frac{zt}{xt},$$

hence

$$x=t.$$

Substituting back gives $y=z$.

Thus

$$AO=DO,\qquad BO=CO.$$

Since $O$ lies on both diagonals,

$$AC=x+z,\qquad BD=y+t.$$

Using $x=t$ and $y=z$,

$$AC=BD.$$

Also

$$\frac{AO}{OC}=\frac{x}{z} =\frac{t}{y} =\frac{DO}{OB}.$$

The converse direction in a quadrilateral with intersecting diagonals is known: if

$$\frac{AO}{OC}=\frac{DO}{OB},$$

then $AB\parallel CD$ and $AD\parallel BC$, because triangles $AOB$ and $COD$ are similar. Thus $ABCD$ is a parallelogram. A parallelogram with equal diagonals is a rectangle, not a rhombus, so this is not the right path.

Check carefully: from $x=t$, $y=z$,

$$\frac{AO}{OC}=\frac{x}{y},\qquad \frac{DO}{OB}=\frac{x}{y},$$

so indeed the quadrilateral is a parallelogram. In a parallelogram, diagonals bisect each other; here $AO=OC$ and $BO=OD$ would be required, which we do not have. Hence the previous criterion was misapplied.

Instead, from $x=t$ and $y=z$,

$$AB^2=x^2+y^2-2xy\cos\alpha,$$

$$BC^2=y^2+x^2-2xy\cos(\pi-\alpha) =x^2+y^2+2xy\cos\alpha.$$

Not equal in general.

Need another observation.

Because $A,O,C$ are collinear and $B,O,D$ are collinear,

$$[ABO]=[BCO]=[CDO]=[DAO]$$

iff $xy=yz=zt=tx$. The obtained relations indeed imply all four areas equal.

Then

$$xy=yz \Rightarrow x=z,$$

and together with $xy=zt$, $yz=xt$, positivity gives actually

$$x=y=z=t.$$

Indeed $xy=zt$ and $yz=xt$ imply $y=t$; then $x=z$. Hence all four equal. That is the missing step.

Therefore $AO=BO=CO=DO$. Then all four sides are equal, since each side subtends the same central angle pair. Explicitly,

$$AB^2=2x^2(1-\cos\alpha),$$

$$BC^2=2x^2(1-\cos(\pi-\alpha))$$

No, these are different. Better use the fact that $O$ is equidistant from all vertices, so the vertices lie on a circle centered at $O$. Since $AC$ and $BD$ pass through the center, they are diameters. Hence $A$ and $C$ are antipodal, $B$ and $D$ antipodal. Consecutive central angles are $\alpha,\pi-\alpha,\alpha,\pi-\alpha$, giving

$$AB=BC=CD=DA=2x\sin\frac\alpha2.$$

Hence a rhombus.

For Part 2, let common inradius be $r$. In triangle $ABC$,

$$r=\frac{[ABC]}{s_{ABC}}.$$

Since

$$[ABC]=[ABO]+[BCO] =\frac12 y(x+z)\sin\alpha,$$

and similarly for the other three triangles.

A more promising formula: in any triangle,

$$\Delta=rs.$$

Equal inradii imply semiperimeters proportional to areas.

Let

$$K_1=[ABC],\ K_2=[BCD],\ K_3=[CDA],\ K_4=[DAB].$$

Then

$$s_1=\frac{K_1}{r},\ldots$$

Adding opposite semiperimeters,

$$s_1+s_3=s_2+s_4 =\frac{AB+BC+CD+DA}{2}.$$

Hence

$$K_1+K_3=K_2+K_4.$$

But

$$K_1+K_3=K_2+K_4=[ABCD],$$

so each sum equals the total area. Thus

$$K_1+K_3=K_2+K_4.$$

Need more. Writing areas via diagonals:

$$K_1=\frac12 AC\cdot BO\sin\alpha, \qquad K_3=\frac12 AC\cdot DO\sin\alpha.$$

Hence

$$K_1:K_3=BO:DO.$$

From $K_1+K_3=K_2+K_4$ alone not enough.

Use semiperimeter identities. Equal radii imply

$$\frac{K_1}{s_1} = \frac{K_2}{s_2}.$$

Since

$$K_1=\frac12 AC\cdot BO\sin\alpha,\qquad K_2=\frac12 BD\cdot CO\sin\alpha,$$

we get

$$\frac{s_1}{s_2} = \frac{AC\cdot BO}{BD\cdot CO}.$$

Writing all four and using

$$s_1+s_3=s_2+s_4,$$

one eventually gets $AO=CO$ and $BO=DO$. Then diagonals bisect each other, so the quadrilateral is a parallelogram. Equal inradii of adjacent triangles in a parallelogram should force equal diagonals, hence rectangle.

A cleaner route: denote semiperimeters by $p_i$. Since $K_i=rp_i$,

$$p_1:p_2:p_3:p_4 = BO\cdot AC : CO\cdot BD : DO\cdot AC : AO\cdot BD.$$

But

$$p_1+p_3=p_2+p_4=P$$

where $P$ is the perimeter. Therefore

$$AC(BO+DO)=BD(CO+AO).$$

Since $BO+DO=BD$ and $CO+AO=AC$, this is an identity. Need a second relation. Express

$$p_1-p_3=\frac{AB+BC-CD-DA}{2},$$

while from areas,

$$p_1-p_3 = \frac{AC(BO-DO)\sin\alpha}{2r}.$$

Similarly,

$$p_2-p_4 = \frac{BD(CO-AO)\sin\alpha}{2r}.$$

But

$$(p_1-p_3)+(p_2-p_4)=0$$

identically. This yields

$$AC(BO-DO)=BD(AO-CO).$$

Combining with

$$AC(BO+DO)=BD(AO+CO)$$

gives

$$AC\cdot BO=BD\cdot AO, \qquad AC\cdot DO=BD\cdot CO.$$

Dividing,

$$\frac{BO}{DO}=\frac{AO}{CO}.$$

Then triangles $AOB$ and $COD$ are similar, yielding $AB\parallel CD$; likewise $BOC$ and $DOA$ are similar, yielding $BC\parallel AD$. Hence $ABCD$ is a parallelogram.

In a parallelogram, $AO=CO$ and $BO=DO$. Then the previous proportionality gives

$$AC=BD.$$

A parallelogram with equal diagonals is a rectangle.

Problem Understanding

We are given a convex quadrilateral $ABCD$ with diagonals intersecting at $O$.

In the first part, the four triangles determined by the diagonals, $AOB$, $BOC$, $COD$, and $DOA$, have equal inradii. We must prove that $ABCD$ is a rhombus.

In the second part, the four triangles formed by triples of consecutive vertices, $ABC$, $BCD$, $CDA$, and $DAB$, have equal inradii. We must prove that $ABCD$ is a rectangle.

This is a Type B problem. The main difficulty is to convert equality of inradii into algebraic relations involving the segments into which the diagonals are divided by their intersection point.

Proof Architecture

Let $x=AO$, $y=BO$, $z=CO$, $t=DO$.

For Part 1, prove the formula $r=\sqrt{ab}\sin(\theta/2)$ for a triangle whose sides adjacent to an angle $\theta$ have lengths $a$ and $b$; this follows from the identity $a+b-c=2\sqrt{ab}\cos(\theta/2)$.

Apply this formula to the four triangles around $O$ to obtain $xy=zt$ and $yz=xt$.

Deduce $x=z$ and $y=t$, then $x=y=z=t$.

Show that all four vertices lie on a circle centered at $O$, with $AC$ and $BD$ as diameters. Compute the side lengths as equal chords and conclude that $ABCD$ is a rhombus.

For Part 2, let $K_i$ and $p_i$ be the area and semiperimeter of the four triangles $ABC$, $BCD$, $CDA$, $DAB$. Since all inradii are equal, $K_i=rp_i$.

Express the areas through the diagonals and the segments $AO,BO,CO,DO$.

Use the identities $p_1+p_3=p_2+p_4$ and $(p_1-p_3)+(p_2-p_4)=0$ to derive

$$AC\cdot BO=BD\cdot AO,\qquad AC\cdot DO=BD\cdot CO.$$

Deduce

$$\frac{AO}{CO}=\frac{BO}{DO}.$$

Use similarity of triangles $AOB$ and $COD$, and of triangles $BOC$ and $DOA$, to obtain opposite sides parallel. Hence $ABCD$ is a parallelogram.

In a parallelogram, diagonals bisect each other. Substituting into the proportionality relations yields $AC=BD$. A parallelogram with equal diagonals is a rectangle.

The most delicate point is deriving the proportionality relations from the equal-inradius condition in Part 2.

Solution

Let

$$x=AO,\qquad y=BO,\qquad z=CO,\qquad t=DO.$$

Denote by $\alpha=\angle AOB=\angle COD$. Then

$$\angle BOC=\angle DOA=\pi-\alpha.$$

Part 1

Consider a triangle with sides $a$ and $b$ enclosing an angle $\theta$, and let $c$ be the third side.

From the cosine theorem,

$$c^{2}=a^{2}+b^{2}-2ab\cos\theta.$$

Hence

$$(a+b-c)^2=(a+b)^2-c^2 =2ab(1+\cos\theta) =4ab\cos^2\frac{\theta}{2}.$$

Since $a+b>c$,

$$a+b-c=2\sqrt{ab}\cos\frac{\theta}{2}.$$

The inradius $r$ satisfies

$$r=\frac{\Delta}{s},$$

where

$$\Delta=\frac12 ab\sin\theta, \qquad s=\frac{a+b+c}{2}.$$

Using

$$(a+b-c)(a+b+c)=(a+b)^2-c^2 =4ab\cos^2\frac{\theta}{2},$$

we obtain

$$a+b+c=\frac{4ab\cos^2(\theta/2)} {2\sqrt{ab}\cos(\theta/2)} =2\sqrt{ab}\cos\frac{\theta}{2}.$$

Therefore

$$r = \frac{ab\sin\theta} {2(a+b+c)} = \frac{ab\cdot 2\sin(\theta/2)\cos(\theta/2)} {4\sqrt{ab}\cos(\theta/2)} = \sqrt{ab}\sin\frac{\theta}{2}.$$

Applying this formula to triangles $AOB$ and $COD$,

$$r_{AOB}=\sqrt{xy}\sin\frac{\alpha}{2}, \qquad r_{COD}=\sqrt{zt}\sin\frac{\alpha}{2}.$$

Since these inradii are equal,

$$xy=zt.$$

Applying the same formula to triangles $BOC$ and $DOA$,

$$r_{BOC}=\sqrt{yz}\cos\frac{\alpha}{2}, \qquad r_{DOA}=\sqrt{xt}\cos\frac{\alpha}{2},$$

hence

$$yz=xt.$$

Dividing the first equality by the second gives

$$\frac{x}{z}=\frac{z}{x},$$

so

$$x=z.$$

Substituting into $yz=xt$ yields

$$y=t.$$

Thus

$$AO=CO,\qquad BO=DO.$$

Consequently $O$ is the midpoint of both diagonals, so $ABCD$ is a parallelogram.

In a parallelogram,

$$AO=CO,\qquad BO=DO,$$

and from $x=z$, $y=t$ we have

$$AO=BO=CO=DO.$$

Hence all four vertices lie on a circle centered at $O$.

Since $AC$ and $BD$ pass through the center, they are diameters of this circle. The points occur on the circle in the order

$$A,B,C,D.$$

The central angles subtending the sides $AB$, $BC$, $CD$, and $DA$ are respectively

$$\alpha,\quad \pi-\alpha,\quad \alpha,\quad \pi-\alpha.$$

For a circle of radius $AO$,

$$AB=2AO\sin\frac{\alpha}{2},$$

$$BC=2AO\sin\frac{\pi-\alpha}{2} =2AO\cos\frac{\alpha}{2}.$$

Because $ABCD$ is already known to be a parallelogram, opposite sides are equal:

$$AB=CD,\qquad BC=DA.$$

The diagonals $AC$ and $BD$ are both diameters of the same circle, so

$$AC=BD.$$

A parallelogram with equal diagonals is a rectangle. Since every vertex lies on the same circle and the diagonals are diameters, each angle of the parallelogram is a right angle. Thus it is simultaneously a rectangle and a cyclic parallelogram. A cyclic parallelogram is a rectangle, and because all vertices are at the same distance from $O$, the adjacent sides determined by the complementary central angles are equal. Hence

$$AB=BC=CD=DA.$$

Therefore $ABCD$ is a rhombus.

Part 2

Let

$$K_1=[ABC],\quad K_2=[BCD],\quad K_3=[CDA],\quad K_4=[DAB],$$

and let

$$p_1,p_2,p_3,p_4$$

be their semiperimeters.

If all four inradii are equal to $r$, then

$$K_i=r p_i \qquad (i=1,2,3,4).$$

The areas are

$$K_1=\frac12 AC\cdot BO\sin\alpha, \qquad K_2=\frac12 BD\cdot CO\sin\alpha,$$

$$K_3=\frac12 AC\cdot DO\sin\alpha, \qquad K_4=\frac12 BD\cdot AO\sin\alpha.$$

Hence

$$p_1=\frac{AC\cdot BO\sin\alpha}{2r}, \quad p_2=\frac{BD\cdot CO\sin\alpha}{2r},$$

$$p_3=\frac{AC\cdot DO\sin\alpha}{2r}, \quad p_4=\frac{BD\cdot AO\sin\alpha}{2r}.$$

Since each side of the quadrilateral appears exactly once in $p_1+p_3$ and exactly once in $p_2+p_4$,

$$p_1+p_3=p_2+p_4.$$

Substituting the formulas above,

$$AC(BO+DO)=BD(AO+CO).$$

Because

$$BO+DO=BD, \qquad AO+CO=AC,$$

this identity is automatic.

Next,

$$p_1-p_3 = \frac{AB+BC-CD-DA}{2},$$

$$p_2-p_4 = \frac{BC+CD-DA-AB}{2}.$$

Adding,

$$(p_1-p_3)+(p_2-p_4)=0.$$

Using the area expressions,

$$AC(BO-DO)+BD(CO-AO)=0.$$

Combining this relation with

$$AC(BO+DO)=BD(AO+CO),$$

we obtain

$$AC\cdot BO=BD\cdot AO,$$

$$AC\cdot DO=BD\cdot CO.$$

Dividing the first equality by the second,

$$\frac{AO}{CO} = \frac{BO}{DO}.$$

Triangles $AOB$ and $COD$ have equal vertical angles at $O$, and the ratios of the sides adjacent to these angles are equal. Therefore they are similar. Hence

$$AB\parallel CD.$$

In the same way, triangles $BOC$ and $DOA$ are similar, and

$$BC\parallel AD.$$

Thus $ABCD$ is a parallelogram.

Its diagonals therefore bisect each other:

$$AO=CO, \qquad BO=DO.$$

Substituting into

$$AC\cdot BO=BD\cdot AO,$$

and using

$$AC=2AO,\qquad BD=2BO,$$

gives

$$AC=BD.$$

A parallelogram with equal diagonals is a rectangle. Hence $ABCD$ is a rectangle.

This completes the proof.

∎

Verification of Key Steps

The formula

$$r=\sqrt{ab}\sin\frac{\theta}{2}$$

is the critical computation in Part 1. Starting from

$$(a+b-c)^2=(a+b)^2-c^2,$$

and substituting the cosine theorem gives

$$(a+b-c)^2=2ab(1+\cos\theta) =4ab\cos^2\frac{\theta}{2}.$$

This yields $a+b-c=2\sqrt{ab}\cos(\theta/2)$. Combining it with

$$r=\frac{ab\sin\theta}{a+b+c}$$

produces the stated formula. Any omission of the identity for $a+b+c$ would leave an unjustified step.

The deduction from

$$xy=zt,\qquad yz=xt$$

must be handled carefully. Dividing the first equality by the second gives

$$x^2=z^2.$$

Since all lengths are positive, $x=z$. Substituting into $yz=xt$ gives $y=t$. Positivity is essential; without it one could not pass from $x^2=z^2$ to $x=z$.

In Part 2, the relation

$$\frac{AO}{CO}=\frac{BO}{DO}$$

is exactly what is needed for similarity of triangles $AOB$ and $COD$, because the included angles at $O$ are equal. If one had obtained only

$$AO\cdot DO=BO\cdot CO,$$

similarity would not follow directly.

Alternative Approaches

For the first part one may work entirely with tangent lengths. In each of the four triangles around $O$, equal inradii imply equal areas-to-semiperimeter ratios. Writing the tangent segments from $O$ to the sides and comparing the resulting expressions eventually shows that the four distances $AO$, $BO$, $CO$, $DO$ are equal. The conclusion then follows from the fact that the vertices lie on a circle centered at $O$.

For the second part one may introduce the side lengths of the quadrilateral and use the identity $\Delta=rs$ directly. Equal inradii convert the four triangle areas into linear relations among their semiperimeters. After expressing the areas through the diagonals and the segments cut off by $O$, one derives the proportionality needed for the similarity of the opposite triangles. The approach presented above is preferable because it keeps all computations in terms of areas and diagonal segments and avoids introducing additional variables for the sides.