Kvant Math Problem 1529
The equalities
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 $$\angle AGB=\angle DHE=120^\circ.\tag{*}$$
- Prove that $$AG+GB+GH+DH+HE\ge CF.$$
- 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.