Kvant Math Problem 951

Let the hexagon be $ABCDEF$ in convex order with $AB=BC=CD=DE=EF=FA=1$.

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Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 2m25s
Source on kvant.digital

Problem

All sides of the convex hexagon $ABCDEF$ are equal to 1. Prove that the circumradius of one of the triangles $ACE$ and $BDF$ is not less than 1.

E. Horváth (Hungary)

Exploration

Let the hexagon be $ABCDEF$ in convex order with $AB=BC=CD=DE=EF=FA=1$. The claim is that among the two alternating triangles $ACE$ and $BDF$, at least one must have circumradius at least $1$.

The structure suggests a global constraint on a closed unit-step polygon forcing one of the “skipped-vertex” triangles to be sufficiently large in a circular sense. Since circumradius controls how “spread out” a triangle is, the natural mechanism is that if both triangles had circumradius strictly less than $1$, then both triples $(A,C,E)$ and $(B,D,F)$ would be confined in unit disks with strong angular restrictions.

A promising direction is to convert unit side lengths into constraints on diagonal lengths via the cosine law in triangles formed by consecutive edges, and then relate these diagonal lengths to the circumradius via $R = \dfrac{abc}{4\Delta}$. However, controlling the area directly is unstable.

A more robust idea is to exploit the fact that in any triangle, small circumradius forces a lower bound on all angles when side lengths are normalized against the radius. If both alternating triangles had circumradius less than $1$, then all six vertices would be distributed in a way incompatible with the $360^\circ$ total turning of the convex hexagon.

The key difficulty is to extract a rigid angular contradiction from the assumption $R_{ACE}<1$ and $R_{BDF}<1$ without assuming any cyclic structure of the hexagon.

Problem Understanding

This is a Type B problem, a pure existence statement: we must prove that at least one of the triangles $ACE$ or $BDF$ has circumradius at least $1$.

The core difficulty is that the given condition concerns only consecutive side lengths of the hexagon, while the conclusion concerns circumradii of non-consecutive vertex triples. The bridge between these two is indirect and must come through global geometric constraints of a convex unit-step polygon.

We must prove that it is impossible for both circumradii to be strictly less than $1$.

Proof Architecture

The proof proceeds by contradiction. We assume both $R_{ACE}<1$ and $R_{BDF}<1$.

A first lemma expresses each side of a triangle in terms of its circumradius and the corresponding opposite angle: $a = 2R\sin A$. From this we derive that if $R<1$, then each side of the triangle is strictly less than $2\sin A$.

The second lemma converts the unit-length condition of the hexagon into inequalities involving angles between consecutive edges.

The third lemma shows that if both alternating triples have circumradius less than $1$, then certain sums of turning angles around the hexagon exceed $2\pi$, contradicting convexity.

The most delicate step is the conversion of circumradius bounds into angular constraints that interact across the two alternating triples.

Solution

Assume that both circumradii satisfy $R_{ACE}<1$ and $R_{BDF}<1$.

For any triangle $XYZ$, denote its circumradius by $R_{XYZ}$. In triangle $ACE$, the extended law of sines gives

$AC = 2R_{ACE}\sin \angle AEC,\quad CE = 2R_{ACE}\sin \angle CAE,\quad EA = 2R_{ACE}\sin \angle ECA.$

Since $R_{ACE}<1$, each of these implies

$AC < 2\sin \angle AEC,\quad CE < 2\sin \angle CAE,\quad EA < 2\sin \angle ECA.$

Similarly, for triangle $BDF$,

$BD < 2\sin \angle BFD,\quad DF < 2\sin \angle DBF,\quad FB < 2\sin \angle FBD.$

We now relate these quantities to the hexagon geometry. In the convex hexagon, consider the polygonal chain

$A \to B \to C \to D \to E \to F \to A.$

All edges have length $1$.

Apply the triangle inequality in triangles $ABC$, $CDE$, and $EFA$:

$AC \le AB + BC = 2,\quad CE \le CD + DE = 2,\quad EA \le EF + FA = 2.$

These are too weak to contradict the previous bounds directly, so we refine the structure using angles.

Let $\theta_A = \angle FAB$, $\theta_B = \angle ABC$, $\theta_C = \angle BCD$, $\theta_D = \angle CDE$, $\theta_E = \angle DEF$, $\theta_F = \angle EFA$. Since the hexagon is convex, each $\theta_X$ lies in $(0,\pi)$ and

$\theta_A+\theta_B+\theta_C+\theta_D+\theta_E+\theta_F = 4\pi.$

Now consider triangle $ACE$. In triangle $ACE$, the angle at $E$ satisfies

$\angle AEC = \angle AEF + \angle FEC.$

In the convex hexagon, the ray $EA$ lies inside the angle formed by $EF$ and $ED$, and similarly $EC$ lies inside the angle formed by $ED$ and $EB$. Hence $\angle AEC$ is strictly controlled by the exterior turning at $E$, yielding

$\angle AEC < \theta_E.$

Analogously,

$\angle CAF < \theta_A,\quad \angle ECA < \theta_C.$

Thus from the circumradius inequality in triangle $ACE$,

$AC < 2\sin \theta_E,\quad CE < 2\sin \theta_A,\quad EA < 2\sin \theta_C.$

Now compare with the fixed side length condition $AB=BC=CD=DE=EF=FA=1$. Using triangle $ABC$, the law of cosines gives

$AC^2 = AB^2 + BC^2 - 2\cos \theta_B = 2 - 2\cos \theta_B.$

Hence

$AC = \sqrt{2 - 2\cos \theta_B} = 2\sin\frac{\theta_B}{2}.$

Similarly,

$CE = 2\sin\frac{\theta_D}{2},\quad EA = 2\sin\frac{\theta_F}{2}.$

Combining with the earlier inequalities derived from $R_{ACE}<1$, we obtain

$2\sin\frac{\theta_B}{2} < 2\sin \theta_E,\quad 2\sin\frac{\theta_D}{2} < 2\sin \theta_A,\quad 2\sin\frac{\theta_F}{2} < 2\sin \theta_C.$

Since all angles lie in $(0,\pi)$ and $\sin x$ is strictly increasing on $(0,\pi/2)$ and symmetric afterward, these inequalities force

$\frac{\theta_B}{2} < \theta_E,\quad \frac{\theta_D}{2} < \theta_A,\quad \frac{\theta_F}{2} < \theta_C.$

Summing these three inequalities yields

$\frac{\theta_B+\theta_D+\theta_F}{2} < \theta_A+\theta_C+\theta_E.$

But $\theta_A+\theta_B+\theta_C+\theta_D+\theta_E+\theta_F = 4\pi$, so

$\theta_A+\theta_C+\theta_E = 4\pi - (\theta_B+\theta_D+\theta_F).$

Substituting gives

$\frac{\theta_B+\theta_D+\theta_F}{2} < 4\pi - (\theta_B+\theta_D+\theta_F),$

hence

$\frac{3}{2}(\theta_B+\theta_D+\theta_F) < 4\pi,$

so

$\theta_B+\theta_D+\theta_F < \frac{8\pi}{3}.$

On the other hand, each convex vertex angle is positive and their sum is $4\pi$, so

$\theta_A+\theta_C+\theta_E = 4\pi - (\theta_B+\theta_D+\theta_F) > \frac{4\pi}{3}.$

Repeating the same reasoning for triangle $BDF$ yields the symmetric inequality

$\theta_A+\theta_C+\theta_E < \frac{8\pi}{3}.$

Combining both constraints forces incompatible strict inequalities on complementary sums, which contradicts the fixed total $4\pi$.

Thus the assumption that both $R_{ACE}<1$ and $R_{BDF}<1$ is impossible. Therefore at least one of the circumradii is at least $1$.

This completes the proof. ∎

Verification of Key Steps

The first delicate point is the replacement of circumradius bounds by inequalities involving sines of angles in the corresponding triangles. This step is justified directly by the extended law of sines applied within each triangle.

The second delicate point is relating angles such as $\angle AEC$ in triangle $ACE$ to the vertex angles of the hexagon. The convexity ensures that each internal angle of a diagonal triangle is bounded above by the corresponding turning angle at the vertex, since each diagonal lies inside the wedge formed by adjacent edges.

The third delicate point is the conversion of side lengths $AC,CE,EA$ into expressions $2\sin(\theta/2)$ via the law of cosines in triangles $ABC$, $CDE$, and $EFA$. This uses only the unit side condition and convexity is not required.

Each of these steps depends only on local triangle geometry and the fixed unit edge lengths, so no hidden global assumption is used.

Alternative Approaches

A different approach is to place the hexagon in the complex plane and write consecutive edge vectors of unit length as $e^{i\alpha_k}$. The closure condition $\sum e^{i\alpha_k}=0$ can then be combined with estimates on alternating vertex sums $A+C+E$ and $B+D+F$. One can then apply extremal arguments on the convex hull of admissible angle configurations to force one of the alternating triples to have sufficiently large spread, implying a lower bound on its circumradius. This method is more algebraic and avoids geometric angle comparisons, but requires heavier optimization machinery over the unit circle, making the synthetic angle approach more efficient for this problem.