IMO 1996 SL G5

Origin: ARM | Category: Geometry

Problem

Let $ABCDEF$ be a convex hexagon such that $AB \parallel DE$, $BC \parallel EF$, and $CD \parallel AF$. Let $R_A$, $R_C$, $R_E$ be the circumradii of triangles $FAB$, $BCD$, $DEF$ respectively, and let $P$ denote the perimeter of the hexagon. Prove that

$$R_A + R_C + R_E \geq \frac{P}{2}.$$

Solution

Let $a, b, c, d, e, f$ denote the lengths of the sides $AB, BC, CD, DE, EF,$ and $FA$ respectively. Note that $\angle A = \angle D$, $\angle B = \angle E$, and $\angle C = \angle F$. Draw the lines $PQ$ and $RS$ through $A$ and $D$ perpendicular to $BC$ and $EF$ respectively ($P, R \in BC$, $Q, S \in EF$). Then $BF \geq PQ = RS$. Therefore

$$2BF \geq PQ + RS,$$

or

(Figure: A, B, C, D, Q, E, F, P, R, S, a, b, c, d, e, f)

$$2BF \geq (a \sin B + f \sin C) + (c \sin C + d \sin B),$$

and similarly,

$$2BD \geq (c \sin A + b \sin B) + (e \sin B + f \sin A), \quad 2DF \geq (e \sin C + d \sin A) + (a \sin A + b \sin C). \tag{1}$$

Next, we have the following formulas for the considered circumradii:

$$R_A = \frac{BF}{2 \sin A}, \quad R_C = \frac{BD}{2 \sin C}, \quad R_E = \frac{DF}{2 \sin E}.$$

It follows from (1) that

$$R_A + R_C + R_E \geq \frac{1}{4} \Big(\frac{a \sin B}{\sin A} + \frac{f \sin A}{\sin B} + \frac{c \sin C}{\sin B} + \frac{d \sin B}{\sin C} + \cdots \Big) \geq \frac{1}{2} (a+b+\cdots) = \frac{P}{2},$$

with equality if and only if $\angle A = \angle B = \angle C = 120^\circ$ and $BF \perp BC$ etc., i.e., if and only if the hexagon is regular.

Second solution. Let us construct points $A^{\prime\prime}, C^{\prime\prime}, E^{\prime\prime}$ such that $ABA^{\prime\prime}F$, $CDC^{\prime\prime}B$, and $EFE^{\prime\prime}D$ are parallelograms. It follows that $A^{\prime\prime}, C^{\prime\prime}, B$ are collinear and also $C^{\prime\prime}, E^{\prime\prime}, B$ and $E^{\prime\prime}, A^{\prime\prime}, F$. Furthermore, let $A'$ be the intersection of the perpendiculars through $F$ and $B$ to $FA^{\prime\prime}$ and $BA^{\prime\prime}$, respectively, and let $C'$ and $E'$ be analogously defined. Since $A'FA^{\prime\prime}B$ is cyclic with the diameter being $A'A^{\prime\prime}$ and since $\triangle FA^{\prime\prime}B \sim \triangle BAF$, it follows that

$$2R_A = A'A^{\prime\prime} = x.$$

(Figure: A, B, C, D, E, F, A', C', E', A'', C'', E'')

Similarly,

$$2R_C = C'C^{\prime\prime} = y, \quad 2R_E = E'E^{\prime\prime} = z.$$

We also have

$$AB = FA^{\prime\prime} = ya, \quad AF = A^{\prime\prime}B = za, \quad CD = C^{\prime\prime}B = zc, \quad CB = C^{\prime\prime}D = xc,$$

$$EF = E^{\prime\prime}D = xe, \quad ED = E^{\prime\prime}F = ye.$$

The original inequality we must prove now becomes

$$x + y + z \geq ya + za + zc + xc + xe + ye. \tag{1}$$

We now follow and generalize the standard proof of the Erdős–Mordell inequality (for the triangle $A'C'E'$), which is what (1) is equivalent to when $A^{\prime\prime} = C^{\prime\prime} = E^{\prime\prime}$.

We set $C'E' = a$, $A'E' = c$ and $A'C' = e$. Let $A_1$ be the point symmetric to $A^{\prime\prime}$ with respect to the bisector of $\angle E'A'C'$. Let $F_1$ and $B_1$ be the feet of the perpendiculars from $A_1$ to $A'C'$ and $A'E'$, respectively. In that case, $A_1F_1 = A^{\prime\prime}F = ya$ and $A_1B_1 = A^{\prime\prime}B = za$. We have

$$ax = A'A_1 \cdot E'C' \geq 2S_{A'E'A_1C'} = 2S_{A'E'A_1} + 2S_{A'C'A_1} = cza + eya.$$

Similarly,

$$cy \geq exc + azc, \quad ez \geq aye + cxe.$$

Thus

$$x + y + z \geq c \frac{za + zc}{a} + c \frac{za - zc}{a} + \cdots. \tag{2}$$

Let us set

$$a_1 = xc - xe, \quad c_1 = ye - ya, \quad e_1 = za - zc.$$

We note that $\triangle A^{\prime\prime}C^{\prime\prime}E^{\prime\prime} \sim \triangle A'C'E'$ and hence $a_1/a = c_1/c = e_1/e = k$. Thus

$$\frac{c}{a} - \frac{a}{c} e_1 + \frac{e}{c} - \frac{c}{e} a_1 + \frac{a}{e} - \frac{e}{a} c_1 = k \Big(\frac{ce}{a} - \frac{ae}{c} + \frac{ea}{c} - \frac{ca}{e} + \frac{ac}{e} - \frac{ec}{a} \Big) = 0.$$

Equation (2) reduces to

$$x + y + z \geq \frac{c}{a} + \frac{a}{c} (za + zc) + \frac{e}{c} + \frac{c}{e} (xe + xc) + \frac{a}{e} + \frac{e}{a} (ya + ye).$$

Using the inequalities

$$\frac{c}{a} + \frac{a}{c} \geq 2, \quad \frac{e}{c} + \frac{c}{e} \geq 2, \quad \frac{a}{e} + \frac{e}{a} \geq 2,$$

we finally get

$$x + y + z \geq ya + za + zc + xc + xe + ye.$$

Equality holds if and only if $a = c = e$ and $A^{\prime\prime} = C^{\prime\prime} = E^{\prime\prime}$ is the center of $\triangle A'C'E'$, i.e., if and only if $ABCDEF$ is regular.

Remark. From the second proof it is evident that the Erdős–Mordell inequality is a special case of the problem. If $P_a, P_b, P_c$ are the feet of the perpendiculars from a point $P$ inside $\triangle ABC$ to the sides $BC, CA, AB$, and $P_a P P_b P_c', P_b P P_c P_a', P_c P P_a P_b'$ are parallelograms, we can apply the problem to the hexagon $P_a P_c' P_b P_a' P_c P_b'$ to prove the Erdős–Mordell inequality for $\triangle ABC$ and point $P$.