An equilateral triangle is also equiangular---the 3 interior angles are 180/3 = 60 degrees each.

Let the original triangle be triangle ABC. Given: AB = BC = CA = a.

(That means triangle ABC is equilateral and equiangular.)

P is on AB; Q is on BC; M is on CA.

In Calculus, we can get the shortest PQ if we express PQ as a function of any related variable, like the changing length of PM for example, and then we equate the first derivative of PQ with respect to PM to zero.

Let us do that.

Let again PQ = x, and PM = y, and QM = z

In triangles APM and ABC,

PM is parallel to BC, MA is parallel to CA, so, angle PMA = angle C = 60 degrees.

(Angles whose sides are parallel each to each are congruent.)

Then, angle APM = 180 -60 -60 = 60 degrees.

So, triangle APM is equiangular and equilateral.

Hence, PM (or y) = AM ------------------------------(i)

In triangles CQM and ABC,

QM is parallel to BA, MC is parallel to AC, so, angle QMC = angle A = 60 degrees.

(Angles whose sides are parallel each to each are congruent.)

Then, angle CQM = 180 -60 -60 = 60 degrees.

So, triangle CQM is equiangular and equilateral.

Hence, QM (or z) = MC ------------------------------(ii)

AC = a ---given.

AC = AM +MC = y +Z = a

Or, z = a -y ----------------(iii)

In triangle PMQ,

>>>PQ = x

>>>angle opposite x is angle PMQ. Angle PMQ = 180 -(angle PMA) -(angle QMC) = 180 -60 -60 = 60 degrees.

>>>PM = y

>>>QM = z = (a-y)

By Law of Cosines,

x^2 = y^2 +(a-y)^2 -2(y)(a-y)cos(60deg)

x^2 = y^2 +(a-y)^2 -2(y)(a-y)(1/2)

x^2 = y^2 +(a-y)^2 -(y)(a-y)

x^2 = y^2 +(a-y)^2 -ay +y^2

x^2 = 2y^2 -ay +(a-y)^2 ---------------------(iv)

Differentiate both sides of (iv) with respect to y,

2x(dx/dy) = 2[2y] -a +2(a-y)(-1)

2x(dx/dy) = 4y -a -2a +2y

2x(dx/dy) = 6y -3a --------------------(v)

Set dx/dy to zero,

2x(0) = 6y -3a

0 = 6y -3a

-6y = -3a

y = (-3a)/(-6)

y = a/2 -------------------***

That means, for minimum x or shortest PQ,

y = a/2, or PM = a/2, or AM = a/2.

Or, M is at the midpoint of AC. -------------------answer.

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If M is at the midpoint of AC, then triangle PMQ is also equilateral. Think about that.