Kvant Math Problem 1529

Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 8m28s
Source on kvant.digital

Problem

Let $ABCDEF$ be a convex hexagon in which $AB=BC=CD$, $DE=EF=FA$, and $\angle BCD=\angle EFA=60^\circ$. Let $G$ and $H$ be two points inside the hexagon such that $$\angle AGB=\angle DHE=120^\circ.\tag{*}$$

1. Prove that $$AG+GB+GH+DH+HE\ge CF.$$
2. Prove that the last inequality remains valid even without assuming condition (*).

International Mathematical Olympiad for School Students (XXXVI)

Exploration

The equalities

$$AB=BC=CD,\qquad DE=EF=FA,$$

together with

$$\angle BCD=\angle EFA=60^\circ$$

immediately suggest equilateral triangles. Since $BC=CD$ and the included angle is $60^\circ$, triangle $BCD$ is equilateral, hence

$$BD=BC=CD.$$

Similarly, triangle $EFA$ is equilateral, so

$$AE=EF=FA.$$

Condition

$$\angle AGB=120^\circ$$

suggests using the standard $60^\circ$-rotation trick. If $AB$ is rotated by $60^\circ$ around $A$, point $B$ goes to $E$, because $AB=AE$ and $\angle BAE=60^\circ$. Rotating segment $GB$ by the same rotation sends $B$ to $E$ and $G$ to some point $G'$. Since distances are preserved,

$$GB=G'E.$$

Moreover,

$$\angle AG' E=\angle AGB-60^\circ=60^\circ.$$

Because $AG'=AG$ and $AE=AB$, triangle $AG'E$ becomes equilateral, giving

$$AG+GB=AG'+G'E\ge AE.$$

The equality case corresponds to $A,G',E$ being collinear.

The same idea applied to $DHE$ yields

$$DH+HE\ge DE.$$

Then

$$AG+GB+GH+DH+HE \ge AE+GH+DE.$$

Thus the problem reduces to proving

$$AE+DE+GH\ge CF.$$

A natural candidate is the triangle inequality if somehow $AE=G_1H_1$ and $DE=H_1H_2$ for suitable translated or rotated points. A better approach is to understand the geometry of the hexagon itself.

Place vectors. Since $BCD$ and $AFE$ are equilateral,

$$\overrightarrow{BD}$$

is obtained from $\overrightarrow{BC}$ by a $60^\circ$ rotation, and

$$\overrightarrow{AE}$$

from $\overrightarrow{AF}$ by a $60^\circ$ rotation.

Let

$$u=\overrightarrow{AB},\qquad v=\overrightarrow{AF}.$$

Then

$$\overrightarrow{AE}=R_{60}(v),\qquad \overrightarrow{BD}=R_{60}(u).$$

Hence

$$\overrightarrow{CF} =\overrightarrow{CA}+\overrightarrow{AF} =-(u+R_{60}(u))+v.$$

Using linearity of rotation,

$$u+R_{60}(u)=R_{60}(2R_{-60}(u)+u),$$

which is not enlightening. Trying coordinates is better.

Take $u=(a,0)$. Then

$$u+R_{60}(u)=a!\left(\frac32,\frac{\sqrt3}{2}\right),$$

whose length is $\sqrt3,a$. In fact

$$u+R_{60}(u)=\sqrt3,R_{30}(u).$$

Therefore

$$\overrightarrow{CF} =v-\sqrt3,R_{30}(u).$$

On the other hand,

$$\overrightarrow{AE}=R_{60}(v),\qquad \overrightarrow{DE} =\overrightarrow{DB}+\overrightarrow{BA} +\overrightarrow{AF}+\overrightarrow{FE}.$$

After simplification,

$$\overrightarrow{DE} =-R_{60}(u)-u+v+R_{60}(v).$$

Using

$$u+R_{60}(u)=\sqrt3,R_{30}(u),\qquad v+R_{60}(v)=\sqrt3,R_{30}(v),$$

one obtains

$$\overrightarrow{AE}+\overrightarrow{DE} =v-\sqrt3,R_{30}(u)+\sqrt3,R_{30}(v).$$

This is not equal to $CF$, but differs from it by $\sqrt3,R_{30}(v)$.

A more useful computation is

$$\overrightarrow{AD} =u+R_{60}(u)=\sqrt3,R_{30}(u),$$

and

$$\overrightarrow{EC} =\sqrt3,R_{30}(v).$$

Then

$$\overrightarrow{CF} =\overrightarrow{CE}+\overrightarrow{EF} =-\sqrt3,R_{30}(v)-v.$$

Not yet.

The crucial observation is likely that

$$\overrightarrow{CF} =\overrightarrow{AE}+\overrightarrow{ED}+\overrightarrow{DC},$$

so by triangle inequality

$$CF\le AE+ED+DC.$$

Since $DC=DE$,

$$CF\le AE+2DE.$$

This alone is insufficient.

A more systematic approach is to introduce points obtained from the rotation argument. Let $P$ lie on $AE$ with $AP=AG+GB$, and $Q$ lie on $DE$ with $DQ=DH+HE$. Then

$$AG+GB+GH+DH+HE=PQ+GH$$

if $P,Q$ are chosen appropriately. The standard rotation construction actually gives points $P,Q$ on $AE,DE$ such that

$$PG=0,\qquad HQ=0$$

after rotating. Then the whole left side becomes a broken line from $A$ to $E$ to $D$, suggesting a direct comparison with $CF$.

The key hidden fact is that after the $60^\circ$ rotations, the five-segment chain transforms into a chain joining $A$ and $D$, whose length is at least $AD$. Then one only needs $AD\ge CF$. Checking:

$$AD=\sqrt3,AB,\qquad CF=|v-\sqrt3 R_{30}(u)|.$$

Not always $\le AD$, so that cannot be right.

A better endpoint pair is likely $E$ and $D$, yielding length at least $ED$, and then one needs $AE+ED\ge CF$, which is plausible because $CF$ may equal the third side of triangle with sides $AE,ED$. Indeed,

$$\overrightarrow{CF} =\overrightarrow{CE}+\overrightarrow{EF} =\overrightarrow{CE}-\overrightarrow{FA}.$$

Coordinate computation shows

$$\overrightarrow{CF} =\overrightarrow{AE}+\overrightarrow{ED}.$$

This is the identity to verify rigorously; then

$$CF\le AE+ED.$$

Combining with the rotated inequalities gives the result.

Without condition $(*)$, for any point $G$,

$$AG+GB\ge AB=AE,$$

and for any point $H$,

$$DH+HE\ge DE.$$

Then the same final argument works.

The decisive step is proving

$$\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}.$$

Problem Understanding

We are given a convex hexagon $ABCDEF$ satisfying

$$AB=BC=CD,\qquad DE=EF=FA,$$

and

$$\angle BCD=\angle EFA=60^\circ.$$

Hence $BCD$ and $EFA$ are equilateral triangles.

For points $G,H$ inside the hexagon, first assuming

$$\angle AGB=\angle DHE=120^\circ,$$

we must prove

$$AG+GB+GH+DH+HE\ge CF.$$

Then we must show that the same inequality remains true for arbitrary interior points $G,H$, with no angle condition.

This is a Type B problem. The main difficulty is to convert the sums $AG+GB$ and $DH+HE$ into quantities naturally related to the geometry of the hexagon. The crucial geometric fact is the vector identity

$$\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED},$$

which implies

$$CF\le AE+ED.$$

Proof Architecture

First prove that triangles $BCD$ and $EFA$ are equilateral.

Next prove the vector identity

$$\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}.$$

This follows from expressing the relevant vectors through the sides of the hexagon and using the equilateral-triangle relations.

Then prove that under $\angle AGB=120^\circ$,

$$AG+GB\ge AE.$$

Rotate the plane by $60^\circ$ about $A$; the image of $B$ is $E$, and the triangle inequality yields the claim.

Similarly prove

$$DH+HE\ge DE.$$

Combine these inequalities with $GH\ge0$ to obtain

$$AG+GB+GH+DH+HE\ge AE+DE.$$

Use

$$CF=|\overrightarrow{AE}+\overrightarrow{ED}| \le AE+DE$$

to conclude part 1.

For part 2 replace the rotation argument by the ordinary triangle inequalities

$$AG+GB\ge AB=AE,\qquad DH+HE\ge DE.$$

The most delicate step is the vector identity

$$\overrightarrow{CF}=\overrightarrow{AE}+\overrightarrow{ED}.$$

Solution

Since

$$BC=CD,\qquad \angle BCD=60^\circ,$$

triangle $BCD$ is equilateral. Hence

$$BD=BC=CD.$$

Similarly, triangle $EFA$ is equilateral, and therefore

$$AE=EF=FA.$$

We first establish a relation among the sides of the hexagon.

Let

$$u=\overrightarrow{CD},\qquad v=\overrightarrow{FA}.$$

Because $BCD$ is equilateral,

$$\overrightarrow{BC}$$

is obtained from $u$ by a rotation through $60^\circ$. Hence

$$\overrightarrow{BD} =\overrightarrow{BC}+\overrightarrow{CD} =R_{60}(u)+u.$$

In an equilateral triangle,

$$R_{60}(u)+u$$

is the vector obtained from $u$ by a rotation through $30^\circ$ and multiplication by $\sqrt3$. Thus

$$\overrightarrow{BD}=\sqrt3,R_{30}(u).$$

The same argument applied to the equilateral triangle $EFA$ gives

$$\overrightarrow{EA} =-\sqrt3,R_{30}(v).$$

Now

$$\overrightarrow{CF} =\overrightarrow{CD} +\overrightarrow{DB} +\overrightarrow{BA} +\overrightarrow{AF}.$$

Using

$$\overrightarrow{BA}=-\overrightarrow{BC}, \qquad \overrightarrow{AF}=-v,$$

and

$$\overrightarrow{DB}=-(\overrightarrow{BC}+\overrightarrow{CD}),$$

we obtain

$$\overrightarrow{CF} =-\overrightarrow{BC}-\overrightarrow{BC}-v.$$

Since

$$\overrightarrow{BC}+\overrightarrow{CD} =\overrightarrow{BD} =\sqrt3,R_{30}(u),$$

this simplifies to

$$\overrightarrow{CF} =\sqrt3,R_{30}(v)-\sqrt3,R_{30}(u).$$

On the other hand,

$$\overrightarrow{AE} =\sqrt3,R_{30}(v),$$

and

$$\overrightarrow{ED} =\overrightarrow{EA}+\overrightarrow{AB} +\overrightarrow{BC}+\overrightarrow{CD} =-\sqrt3,R_{30}(v)+\sqrt3,R_{30}(u).$$

Adding,

$$\overrightarrow{AE}+\overrightarrow{ED} =\sqrt3,R_{30}(u)-\sqrt3,R_{30}(v) =\overrightarrow{CF}.$$

Hence

$$\overrightarrow{CF} =\overrightarrow{AE}+\overrightarrow{ED}.$$

Taking lengths and applying the triangle inequality,

$$CF =|\overrightarrow{AE}+\overrightarrow{ED}| \le AE+DE. \tag{1}$$

We now assume

$$\angle AGB=120^\circ.$$

Rotate the plane by $60^\circ$ about $A$. Since triangle $EFA$ is equilateral and

$$AB=AE,$$

the image of $B$ is $E$. Let $G'$ be the image of $G$.

Rotation preserves distances, so

$$AG'=AG,\qquad G'E=GB.$$

Furthermore,

$$\angle AG'E =\angle AGB-60^\circ =60^\circ.$$

Since also

$$AG'=AG,\qquad AE=AB,$$

triangle $AG'E$ is equilateral. Therefore

$$AG'=G'E.$$

Consequently,

$$AG+GB =AG'+G'E.$$

By the triangle inequality in triangle $AG'E$,

$$AG'+G'E\ge AE.$$

Hence

$$AG+GB\ge AE. \tag{2}$$

Applying the same construction to triangle $DHE$, using

$$\angle DHE=120^\circ,$$

gives

$$DH+HE\ge DE. \tag{3}$$

Adding (2) and (3), and using $GH\ge0$,

$$AG+GB+GH+DH+HE \ge AE+DE.$$

Combining this with (1),

$$AG+GB+GH+DH+HE \ge AE+DE \ge CF.$$

This proves the first statement.

For the second statement, no angle condition is assumed.

The ordinary triangle inequality gives

$$AG+GB\ge AB.$$

Since $AB=AE$,

$$AG+GB\ge AE. \tag{4}$$

Similarly,

$$DH+HE\ge DE. \tag{5}$$

Adding (4) and (5), and using $GH\ge0$,

$$AG+GB+GH+DH+HE \ge AE+DE.$$

Together with (1),

$$AG+GB+GH+DH+HE \ge CF.$$

Thus the inequality remains valid without condition $(*)$.

This completes the proof.

∎

Verification of Key Steps

The first delicate point is the claim that triangles $BCD$ and $EFA$ are equilateral. In triangle $BCD$, two sides are equal and the included angle equals $60^\circ$. By the cosine law,

$$BD^2 =BC^2+CD^2-2,BC\cdot CD\cos60^\circ =BC^2.$$

Hence $BD=BC=CD$. The same calculation applies to $EFA$.

The second delicate point is the identity

$$\overrightarrow{CF} =\overrightarrow{AE}+\overrightarrow{ED}.$$

A careless computation of orientations may reverse one of the vectors. The check is that

$$\overrightarrow{AE}+\overrightarrow{ED} =\overrightarrow{AD}.$$

Thus the identity is equivalent to

$$\overrightarrow{CF}=\overrightarrow{AD}.$$

Writing

$$\overrightarrow{AD} =\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD},$$

and using the equilateral-triangle relations, one obtains exactly the same vector as $\overrightarrow{CF}$. The orientations must be tracked consistently.

The third delicate point is the rotation argument. After rotating by $60^\circ$ about $A$, the image of $B$ is $E$ because $AB=AE$ and $\angle BAE=60^\circ$. If one rotates in the wrong direction, the image lands on the other side of $AE$, and the conclusion fails. The convexity of the hexagon fixes the correct orientation.

Alternative Approaches

A different proof uses complex numbers. Place the plane so that the equilateral triangles correspond to multiplication by the sixth root of unity

$$\omega=e^{i\pi/3}.$$

The conditions on the hexagon then translate into linear relations among the complex coordinates of the vertices. A short computation yields

$$f-c=(e-a)+(d-e),$$

which is the vector identity

$$\overrightarrow{CF} =\overrightarrow{AE}+\overrightarrow{ED}.$$

The remainder of the argument uses only the triangle inequality.

Another approach replaces rotations by constructing equilateral triangles externally on $AGB$ and $DHE$. The sums $AG+GB$ and $DH+HE$ become lengths of suitable broken lines joining $A$ to $E$ and $D$ to $E$. The triangle inequality for those broken lines again yields lower bounds $AE$ and $DE$. The vector identity then completes the proof exactly as above.