# A triangle with vertices on the sides of a square, with one at a midpoint, cannot be equilateral

Let A, B, C, D, L, M, N be distinct points in the plane such that A, B, C, D are the vertices of a square with sides AB, BC, CD, DA and L, M, N lie on the sides AB, CD, BC respectively. If M is the mid-point of CD, then show that △LMN is not an equilateral triangle.

I cannot get anywhere with it. I tried to find a contradiction by assuming 60 degree angles, but no luck.

Source: Geometry proof contest math 2b

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Hint: Consider a circle with center $M$ and variable radius. How can it cut $AB$ and $BC$ at the same time?

Hence, show that $\angle LMN \neq 60^\circ$.

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So far, in the case where the circle has radius equal to AB, angle NMC is 30. Now the case where the radius is greater than AB remains, which I will try to show in a bit. – Yadnarav3 Sep 9 '13 at 0:18

There must be a less roundabout solution, but here's one attempt to show it:

Lemma: If an isosceles triangle PAB, with equal angles $15^\circ$ at the ends of its base AB, is drawn inside a square ABCD, as in the diagram below (that includes proof), then the points P, C, D are the vertices of an equilateral triangle.

The bisector of $\angle DPC$ perpendicularly bisects AB (let it be at M). Then an inscribed equilateral triangle with a vertex at M cannot have a vertex on the opposite side DC of the square because translating $\Delta PDC$ by PM to form $\Delta D'MC'$ creates side $D'C' \parallel DC$.

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This is going to be a proof by contradiction, we'll assume that △LMN is equilateral and will try to contradict that.

First we gonna draw a perpendicular line to the side AB, cutting it at point L, like it's done on the picture. Now we gonna apply the Pythagorean Theorem to the right triangles, but before we gonna introduce some notations.

Let: $a=AB=BC=CD=AD$, $x=BN$ and $y=BL$. Now we apply the Pythagorean Theroem.

From △MCN we have:

$$MN^2 = CM^2 + CN^2 = \left(\frac{a}{2}\right)^2 + (a-x)^2$$ $$MN^2 = \frac{a^2}{4} + a^2 - 2ax + x^2$$ $$MN^2 = \frac{5a^2}{4} - 2ax + x^2$$

From △LBN we have:

$$LN^2 = BL^2 + BN^2$$ $$LN^2 = x^2 + y^2$$

From △LEM we have:

$$LM^2 = LE^2 + EM^2 = a^2 + (y-\frac{a}{2})^2$$ $$LM^2 = a^2 + y^2 - ay + \frac{a^2}{4}$$ $$LM^2 = \frac{5a^2}{4} - ay + y^2$$

Because △LMN is equilateral it means that the all sides are equal so we get:

$$MN^2 = LN^2$$ $$\frac{5a^2}{4} - 2ax + x^2 = x^2 + y^2$$ $$\frac{5a^2}{4} - 2ax - y^2 = 0$$

For the other side we have:

$$LM^2 = LN^2$$ $$\frac{5a^2}{4} - ay + y^2 = x^2 + y^2$$ $$\frac{5a^2}{4} - ay - x^2 = 0 \implies y = \frac{\frac{5a^2}{4} - x^2}{a}$$

Note that $a \neq 0$, because $A,B,C,D$ are distinct points. Now we substitute into the previous equation and we get:

$$\frac{5a^2}{4} - 2ax - y^2 = 0$$ $$\frac{5a^2}{4} - 2ax - \left(\frac{\frac{5a^2}{4} - x^2}{a}\right)^2 = 0$$

Little playing and we can't transform this equation to:

$$-\frac{(4x^2 - 8ax + 5a^2)(4x^2 + 8ax + a^2)}{16a^2} = 0$$

Obviously the denominator can't be 0, so we need to find the root of the two quadratic equation. The first one produces 2 solution:

$$x_{1/2} = (1\pm\frac{i}{2})a$$

But $x$ can't have complex number value. The second one produces also 2 solution and they are:

$$x_{3/4} = \frac{a(-2 \pm \sqrt{3})}{2}$$

But this provides that $x$ has negative value, but one side can't have a negative length.

This means that the assumption that △LMN is equilateral is wrong.

Q.E.D.

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