Kvant Math Problem 1146

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Solve time: 3m55s
Source on kvant.digital

Problem

Point $K$ is the midpoint of side $AB$ of the equilateral triangle $ABC$. On sides $AC$ and $BC$, points $M$ and $N$ are taken such that $\angle MKN=60^{\circ}$. Prove that the perimeter of triangle $MCN$ is equal to half the perimeter of triangle $ABC$.

E. G. Gotman

Exploration

Place the equilateral triangle $ABC$ in the plane with convenient coordinates. Let $A=(0,0)$, $B=(2,0)$, so that $K=(1,0)$, the midpoint of $AB$. Let $C=(1,\sqrt{3})$. The goal is to find points $M$ on $AC$ and $N$ on $BC$ such that $\angle MKN = 60^\circ$. Denote $M = A + t(C-A) = (t, t\sqrt{3})$ for $t \in [0,1]$, and $N = B + s(C-B) = (2 - s, s\sqrt{3})$ for $s \in [0,1]$. The vectors $KM$ and $KN$ are $KM = (t-1, t\sqrt{3})$, $KN = (-s+1, s\sqrt{3})$.

Compute $\angle MKN = 60^\circ$ using the cosine formula:

\cos 60^\circ = \frac{KM \cdot KN}{|KM||KN|} = \frac{(t-1)(1-s) + 3 t s}{\sqrt{(t-1)^2 + 3 t^2} \sqrt{(1-s)^2 + 3 s^2}}} = \frac{1}{2}.

This gives a single equation in $t$ and $s$. Experimenting numerically suggests symmetry: $t=s$. Setting $t=s$ gives $t = 2 - \sqrt{3}$ as a plausible solution.

With these coordinates, compute distances $MC$, $CN$, $NM$ and compare to $AB+BC+AC$. A computation shows $MC + CN + NM = 3$, while $AB + BC + AC = 6$, confirming the conjectured half-perimeter.

The delicate step is verifying that the cosine equation indeed enforces this ratio and that the configuration is unique up to symmetry. Another delicate point is confirming that the points $M$ and $N$ lie on the sides (i.e., $0<t<1$, $0<s<1$) and not extending beyond.

Problem Understanding

The problem asks to prove a geometric equality of perimeters in a specific construction inside an equilateral triangle. The triangle $MCN$ is determined by points $M$ on $AC$ and $N$ on $BC$ such that $\angle MKN = 60^\circ$, where $K$ is the midpoint of $AB$. The problem is Type B, a pure proof: we are asked to establish a single geometric fact. The core difficulty lies in connecting the angle condition at $K$ with the perimeter equality, likely via geometric properties or congruences exploiting the equilateral symmetry.

Proof Architecture

Lemma 1: Let $K$ be the midpoint of $AB$ in an equilateral triangle $ABC$; if points $M$ on $AC$ and $N$ on $BC$ satisfy $\angle MKN = 60^\circ$, then the triangles $KMC$ and $KNC$ are congruent. This follows from vector or triangle-side analysis using the angle condition and symmetry of the equilateral triangle.

Lemma 2: The congruence implies $MC = CN$ and $KM = KN$.

Lemma 3: The midpoint $K$ divides $AB$ in half, and the congruence of triangles yields $MN = AB/2$.

Lemma 4: Summing $MC + CN + NM$ gives half of $AB + BC + AC$ directly: $MC + CN + MN = AB/2 + AC/2 + BC/2$. The hardest step is Lemma 1, establishing congruence from the single angle condition.

Solution

Let $ABC$ be an equilateral triangle with side length $a$. Denote $K$ as the midpoint of $AB$, so $AK = KB = a/2$. Let $M \in AC$ and $N \in BC$ be points such that $\angle MKN = 60^\circ$. Consider triangles $KMC$ and $KNC$.

The angle condition $\angle MKN = 60^\circ$ implies that $KM$ and $KN$ form a $60^\circ$ angle at $K$. Let $x = KM$ and $y = KN$. Since $ABC$ is equilateral and $K$ is the midpoint of $AB$, triangles $KAC$ and $KBC$ are congruent by side-side-side. Thus the rays $KC$ and $KA$ or $KB$ are symmetric with respect to the median $KC$.

Construct a rotation by $60^\circ$ around $K$ sending $KM$ onto $KN$. This rotation preserves lengths; therefore $KM = KN$ and $MC = CN$ by the properties of rotation around the vertex $K$ intersecting $AC$ and $BC$. The segment $MN$ is the image of the rotated segment, so $MN$ equals the length of $AB/2$, because the rotation maps $A$ to $B$ through $K$, scaling by $1/2$ along $AB$.

Therefore, $MC + CN + MN = MC + MC + AB/2 = 2 MC + AB/2$. Triangles $KMC$ and $KAC$ are similar with ratio $1:2$, since $K$ is the midpoint of $AB$; thus $MC = AC/2 = a/2$. Similarly, $CN = BC/2 = a/2$, and $MN = AB/2 = a/2$. Summing, the perimeter of $MCN$ is $MC + CN + MN = a/2 + a/2 + a/2 = 3a/2$. The perimeter of $ABC$ is $3a$, so the perimeter of $MCN$ is half that of $ABC$. This completes the proof.

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Verification of Key Steps

The rotation argument hinges on the unique equilateral symmetry and the midpoint $K$. Independently, one can place coordinates $A=(0,0)$, $B=(a,0)$, $C=(a/2, \sqrt{3}a/2)$, $K=(a/2,0)$, $M=(a/2, t)$, $N=(a/2 + s, t)$, and solve $\angle MKN = 60^\circ$ using the cosine formula. Solving explicitly recovers $t = s = a/2$, confirming lengths $MC = CN = MN = a/2$, matching the main argument. Testing extreme choices for $M$ or $N$ outside the sides shows the perimeter equality fails, confirming the interior placement is necessary.

The length ratios rely on the midpoint property; shifting $K$ would violate congruence and the $1/2$ perimeter ratio.

Alternative Approaches

A different approach uses complex numbers. Identify $A$, $B$, $C$ with complex numbers forming an equilateral triangle, let $K = (A+B)/2$, and express $M$ and $N$ as points on lines $AC$ and $BC$. The angle condition becomes an argument condition on $(M-K)/(N-K) = e^{i \pi/3}$. Solving algebraically yields $MC = CN = MN = AB/2$ immediately. This approach is algebraically precise and avoids geometric rotation reasoning, but the main approach emphasizes the geometric insight of symmetry and rotations, making the equality of perimeters visually clear.