Kvant Math Problem 701
Let the sides of the acute triangle $LMN$ be
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Problem
Lyuda, Marina, and Natasha drew an acute triangle $LMN$. Then Lyuda constructed her own triangle in which the lengths of two sides are $|LM|$ and $|LN|$, and the angle between them is $60^\circ$ greater than the angle $L$ of triangle $LMN$. Similarly, Marina constructed her own triangle with sides of lengths $|ML|$ and $|MN|$, the angle between which is $60^\circ$ greater than $\angle M$, and Natasha constructed hers in which the angle between the sides $|NL|$ and $|NM|$ is equal to $\angle N + 60^\circ$. Prove that the third (new) sides of the triangles constructed by all three girls are equal.
A. Kaplan
Exploration
Let the sides of the acute triangle $LMN$ be
$$a=|MN|,\qquad b=|NL|,\qquad c=|LM|,$$
opposite the angles
$$L=\alpha,\qquad M=\beta,\qquad N=\gamma.$$
The three newly constructed triangles have third sides, say,
$$x_L,\quad x_M,\quad x_N.$$
The law of cosines immediately gives
$$x_L^2=b^2+c^2-2bc\cos(\alpha+60^\circ),$$
and analogous formulas for $x_M,x_N$.
The problem asks us to show that these three quantities are equal.
Expanding
$$\cos(\alpha+60^\circ) =\frac12\cos\alpha-\frac{\sqrt3}{2}\sin\alpha,$$
yields
$$x_L^2=b^2+c^2-bc\cos\alpha+\sqrt3,bc\sin\alpha.$$
Using the law of cosines in the original triangle,
$$a^2=b^2+c^2-2bc\cos\alpha,$$
hence
$$x_L^2 =a^2+bc\cos\alpha+\sqrt3,bc\sin\alpha.$$
Now
$$bc\sin\alpha=2\Delta,$$
where $\Delta$ is the area of $LMN$, and
$$bc\cos\alpha=\frac{b^2+c^2-a^2}{2}.$$
Substituting,
$$x_L^2 =\frac{a^2+b^2+c^2}{2}+2\sqrt3,\Delta.$$
The right-hand side is completely symmetric in the three sides. If the analogous computation for $x_M^2$ and $x_N^2$ gives the same expression, the claim follows.
The only potentially delicate step is ensuring that the symmetric expression is obtained correctly; a sign error in the trigonometric expansion would destroy the argument.
Problem Understanding
We are given an acute triangle $LMN$. For each vertex, a new triangle is constructed using the two sides adjacent to that vertex, while increasing the included angle by $60^\circ$. The quantity of interest is the third side of each new triangle.
This is a Type B problem: prove a stated property.
The core difficulty is to show that three seemingly different law-of-cosines expressions actually coincide. The natural route is to express each new side through the sides and area of the original triangle and obtain a symmetric formula.
Proof Architecture
First, compute the square of Lyuda's new side by the law of cosines in her constructed triangle.
Second, rewrite the resulting expression using the law of cosines in the original triangle and the area formula $\Delta=\frac12 bc\sin\alpha$.
Third, show that
$$x_L^2=\frac{a^2+b^2+c^2}{2}+2\sqrt3,\Delta.$$
Fourth, repeat the same calculation for Marina's and Natasha's sides, obtaining exactly the same symmetric expression.
The lemma most likely to fail under scrutiny is the derivation of the symmetric formula for $x_L^2$; every later step depends on it.
Solution
Let
$$a=|MN|,\qquad b=|NL|,\qquad c=|LM|,$$
and let
$$\alpha=\angle L,\qquad \beta=\angle M,\qquad \gamma=\angle N.$$
Denote by $x_L$, $x_M$, and $x_N$ the third sides of the triangles constructed by Lyuda, Marina, and Natasha respectively.
For Lyuda's triangle, the sides adjacent to the included angle have lengths $b$ and $c$, and the included angle equals $\alpha+60^\circ$. By the law of cosines,
$$x_L^2=b^2+c^2-2bc\cos(\alpha+60^\circ).$$
Using
$$\cos(\alpha+60^\circ) =\frac12\cos\alpha-\frac{\sqrt3}{2}\sin\alpha,$$
we obtain
$$x_L^2 =b^2+c^2-bc\cos\alpha+\sqrt3,bc\sin\alpha.$$
The law of cosines in the original triangle gives
$$a^2=b^2+c^2-2bc\cos\alpha,$$
hence
$$bc\cos\alpha=\frac{b^2+c^2-a^2}{2}.$$
Let $\Delta$ be the area of triangle $LMN$. Then
$$\Delta=\frac12 bc\sin\alpha,$$
so
$$bc\sin\alpha=2\Delta.$$
Substituting these relations into the formula for $x_L^2$,
\begin{align*}
x_L^2
&=b^2+c^2-\frac{b^2+c^2-a^2}{2}+2\sqrt3,\Delta\
&=\frac{a^2+b^2+c^2}{2}+2\sqrt3,\Delta.
\end{align*}
Now perform the same computation for Marina's triangle. Its adjacent sides have lengths $a$ and $c$, and the included angle is $\beta+60^\circ$. Therefore
$$x_M^2 =a^2+c^2-2ac\cos(\beta+60^\circ).$$
Repeating the previous transformations,
$$x_M^2 =\frac{a^2+b^2+c^2}{2}+2\sqrt3,\Delta.$$
Likewise, Natasha's triangle has adjacent sides $a$ and $b$ and included angle $\gamma+60^\circ$, giving
$$x_N^2 =a^2+b^2-2ab\cos(\gamma+60^\circ) =\frac{a^2+b^2+c^2}{2}+2\sqrt3,\Delta.$$
Consequently,
$$x_L^2=x_M^2=x_N^2.$$
All three quantities are lengths and therefore positive, so equality of their squares implies
$$x_L=x_M=x_N.$$
Thus the third sides of the three constructed triangles are equal.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the expansion of $\cos(\alpha+60^\circ)$. Using the addition formula,
$$\cos(\alpha+60^\circ) =\cos\alpha\cos60^\circ-\sin\alpha\sin60^\circ =\frac12\cos\alpha-\frac{\sqrt3}{2}\sin\alpha.$$
Multiplying by $-2bc$ gives
$$-bc\cos\alpha+\sqrt3,bc\sin\alpha,$$
so the sign of the area term is positive.
The second delicate step is the substitution
$$bc\cos\alpha=\frac{b^2+c^2-a^2}{2}.$$
This follows directly from
$$a^2=b^2+c^2-2bc\cos\alpha.$$
Solving for $bc\cos\alpha$ yields exactly the displayed formula.
The final step is passing from equality of squares to equality of lengths. Since $x_L,x_M,x_N$ are side lengths of triangles, each is strictly positive. Therefore
$$x_L^2=x_M^2=x_N^2$$
forces
$$x_L=x_M=x_N.$$
No sign ambiguity remains.
Alternative Approaches
A more geometric proof uses vectors or complex numbers. If the side vectors of triangle $LMN$ are rotated by $60^\circ$, the law of cosines computation can be interpreted as the squared length of a sum of vectors. The resulting expression becomes
$$\frac{a^2+b^2+c^2}{2}+2\sqrt3,\Delta,$$
which is invariant under permutation of the vertices, immediately implying equality of the three new sides.
The trigonometric approach is preferable here because the problem is formulated entirely in terms of side lengths and angles. A single application of the law of cosines and the area formula produces the required symmetric expression with minimal machinery.