58th Ukrainian National Mathematical Olympiad

Question

On sides AB, BC, AC of triangle ABC with BAC = 120° there are points M, K, N respectively so, that MKN is equilateral, and AM = 2,017, AN = 2,018. Baron Munchausen assures, that MKN has the smallest perimeter of all equilateral triangles, that have exactly one vertex on each of sides of ABC. Is baron right?

Accepted Answer

According to the condition, all the angles of

△ M K N

equal to

60°

(Fig. 39). As

∠ B A C + ∠ M K N = 180°

, quadrilateral

A M K N

is cyclic, so

∠ K A C = ∠ K M N = 60° = ∠ M N K = ∠ B A K .

Let's consider points

M_{1}

and

N_{1}

– projections of point

K

on sides

A B

and

A C

respectively. So

△ A K M_{1} = △ A K N_{1}

, so

A M_{1} = A N_{1}

. Apart from that,

∠ M_{1} K N_{1} = 60°

and

K M_{1} = K N_{1}

. So,

△ K M_{1} N_{1}

is also equilateral. Case

N = N_{1}

,

M = M_{1}

is impossible, as otherwise we would have

A M = A N

. So

K M_{1} < K M

, so perimeter of

△ K M_{1} N_{1}

is strictly less than perimeter of

△ K M N

. So, baron Munchausen is not right.