ABCD is a square and CMN is an equilateral triangle, with M on AD and N on AB. If the area of ABCD is one square inch, compute the area of CMN in square inches.
I have been able to find all of the angle values after discovering that triangle MDC and NBC are congruent. This means that triangle AMN is isosceles, and from there one can find all the angle values. However, I have not been able to find anything that will help me find the area. Setting algebraic values and then bashing them has not seemed to work. What is the proper way to approach this question?

  • $\begingroup$ I've given you a non trig solution, in case you're interested. $\endgroup$ – Deepak Oct 19 '17 at 0:13

If the side of an equilateral triangle is $A$, then the altitude is $\sqrt {A^2 - (A/2)^2} = \sqrt {(3/4)A^2} = \frac {\sqrt 3}2 A$

The side of the equilateral triangle here is $\dfrac{S}{\cos 15°}$, with $S$ the length of the square side.

The $15°$ comes from considering $\angle DCM + \angle MCN + \angle NCB = \angle DCB$ (and $\angle DCM=\angle BCN $).

  • 2
    $\begingroup$ This answer is wrong. The right angle in the right triangle with the altitude is opposite the altitude. The method is right, though, so you can fix it. $\endgroup$ – Ethan Bolker Oct 19 '17 at 0:18
  • $\begingroup$ Ah yes. I thought the numbers looked screwy. $\endgroup$ – Joffan Oct 19 '17 at 0:35

Here's a non trigonometric way. Uses Pythagorean Theorem ("PT")and a little algebra.

Let $\displaystyle AM = AN = x$ (you know that $AM$ and $AN$ are equal by symmetry).

Then $\displaystyle BN = DM = 1-x$.

By PT, $\displaystyle MN = \sqrt 2 x$, and thus $\displaystyle CM = CN = \sqrt 2 x$

Now considering the right triangles $CMD$ and $CNB$ and again applying PT, we get:

$$(1-x)^2 + 1^2 = (\sqrt 2 x)^2$$

which quickly reduces to the quadratic:

$$x^2 +2x - 2= 0$$

for which the only permissible solution is $x = \sqrt 3 - 1$

From that we can get the area of $AMN$ as $\displaystyle \frac 12 (\sqrt 3 - 1)^2 = 2-\sqrt 3$

The areas of $CMD$ and $CNB$ add up to $\displaystyle (2)(\frac 12)(1-x) = (1-\sqrt 3 + 1) = 2- \sqrt 3$ (i.e. the combined areas of $CMD$ and $CNB$ are equal to that of $AMN$, an interesting geometric insight).

So the area of the equilateral triangle is simply $\displaystyle 1 - 2(2-\sqrt 3) = 2\sqrt 3 - 3 \ \mathrm {sq. inch.}$

  • $\begingroup$ Would this be equivalent to @Joffan's answer? $\endgroup$ – Gerard L. Oct 19 '17 at 0:13
  • $\begingroup$ Equivalent? It definitely gives the same answer (through a non trig approach), although Joffan did not compute that answer. $\endgroup$ – Deepak Oct 19 '17 at 0:14

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$$\text{area of $2\times 2$ square} \;=\; 4 \;=\; 2\color{purple}{x} \;+\; \color{red}{\frac{1}{2}\cdot 2\cdot\sqrt{3}} \;+\; 5\cdot\color{blue}{\frac{1}{2}\cdot 2\cdot\left(2-\sqrt{3}\right)}$$ $$\to\quad \color{purple}{x} = 2\sqrt{3} - 3$$


Take $C$ as the origin, $CD$ along positive $X$ axis and $CB$ along positive $Y$ axis. Let $CM$ make an angle $\theta$ with the $X$ axis. Then $M$ is $(a, a\sin\theta)$, where $a$ is the side of the square, and $CN$ makes an angle $30^\circ-\theta$ with the $Y$ axis. Length of $CM^2$ is $a^2+a^2\sin^2\theta$ and that of $CN^2$ is $a^2+a^2\sin^2(30^\circ-\theta)$. Equating these, we obtain $\theta = 15^\circ$. Hence the length of $CM^2$ is $a^2 + a^2\sin^2 15^\circ = a^2\left(1+\frac{1-\cos 30^\circ}{2}\right) = a^2\left(\frac{3-\sqrt{3}}{4}\right)$. Now the area of the triangle is $\frac{\sqrt{3}}{4} CM^2$

  • $\begingroup$ Is this solvable without using trig? I feel that there is a much simpler answer. $\endgroup$ – Gerard L. Oct 18 '17 at 23:49
  • $\begingroup$ You can deduce that the angle made by $CM$ with the side is $15^\circ$ as follows: Reflect the figure in the diagonal $CA$. Then by symmetry, the side $CM$ must go to the side $CN$ and hence the angle made by $CM$ with $X$ axis is the same as the angle made by $CN$ with $Y$ axis. It follows that the angle is $15^\circ$ $\endgroup$ – user348749 Oct 18 '17 at 23:52
  • $\begingroup$ Then how would this connect to your trig answer? $\endgroup$ – Gerard L. Oct 18 '17 at 23:53
  • $\begingroup$ You can now compute $CM$. $\endgroup$ – user348749 Oct 18 '17 at 23:54

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