# Finding a missing angle in the picture containing regular hexagon and square

I want to find $$\angle AGM=\theta$$ in the following picture:
Here $$ABCDEF$$ and $$BAGH$$ are regular hexagon and square respectively and $$M$$ is the midpoint of $$FH$$.

I found a trigonometric solution. I'm providing key ideas of the solution:

Let $$AB=1$$. Now we can apply cosine rule on $$\triangle AHF$$ to find $$HF$$ and $$HM$$. Now in $$\triangle MGH$$, we can find $$GM$$ using cosine rule again and then find $$\angle MGH$$ by sine rule. This gives $$\theta=15^{\circ}$$. (I'm not providing the calculations as they are not nice and I did most of them with calculator.)

But I believe there are some beautiful synthetic solution to the but didn't find one. So, I need a synthetic solution to the problem.

• A similar issue : gogeometry.blogspot.com/2011/04/… Commented Aug 25, 2021 at 16:53
• Hint: If $O$ is the center of the hexagon, then $\square OFGH$ is a parallelogram.
– Blue
Commented Aug 25, 2021 at 17:50

Let $$I$$ be a center of hexagon. Then $$HG = ID$$ and they are parallel, so $$IDGH$$ is a paralelogram so $$K$$ is also the midpoint of $$GI$$, thus $$G,K,I$$ are collinear.

Since $$GEI$$ is isosceles triangle and $$\angle GEI = 150^{\circ}$$ we have $$\theta = 15^{\circ}$$.

• Not doubting your answer, just requesting clarification: how did you get $\angle GEI = 150^{\circ}$? Commented Aug 26, 2021 at 16:01
• 90+60 .................. @blackbrandt Commented Aug 26, 2021 at 16:07
• wow I actually feel dumb now :) i was totally overthinking that Commented Aug 26, 2021 at 16:39
• The statement that HG = ID is implicitly based on the fact that the length of a side of a regular hexagon is equal to the length from the center of the hexagon to any vertex, I believe. I think you should make that explicit; it took me only a few moments to realize that’s how you got that conclusion and convince myself that it was true, but it’s better for such things to be explicitly stated. Also, it’s a little confusing that you relabeled the points in question; your K is M in the question, your E is A, and your D is F. Makes things a little harder to follow. Commented Aug 26, 2021 at 21:20
• What are you talking? @KRyan That is what kids in elementary school learn/know. Commented Aug 27, 2021 at 11:19

Even without pure geometry, the work can be simplified. Also, see my edit at the end for a synthetic solution.

$$\angle PBC = \angle AFQ = 30^\circ$$

If side length is $$a$$, $$PC = GQ = \frac{a}{2}$$

So, $$PF = HQ = \frac{3a}{2}$$

Similarly, $$PH = FQ = a + \frac{a \sqrt3}{2}$$

Given $$M$$ is the midpoint of $$FH$$,

$$MN = a - \frac{HQ}{2} = \frac{a}{4}$$

$$GN = \frac{FQ}{2} = \frac{a (2 + \sqrt3)}{4}$$

$$\tan \theta = \frac{1}{2 + \sqrt3} = 2 - \sqrt3 \implies \theta = 15^0$$

Synthetic solution (using similar construct as above):

Given $$M$$ is the midpoint of $$HF$$, it is also the center of rectangle $$HPFQ$$ and hence of rectangle $$GIJK$$.

Also note $$J$$ is the center of the hexagon.

So, $$\triangle FAJ$$ is an equilateral triangle.

$$\angle JAK = 30^\circ$$.

In $$\triangle GAJ$$, $$AG = AJ$$

$$\therefore \angle AGM = \angle AJM = 15^\circ$$

• (+1) This seems extremely close to a synthetic solution. In fact, also using synthetic methods, we can show that in the 15-75-90 triangle ($\triangle GMA$), the length of the altitude from the right angled vertex is 4 times the length of the hypotenuse (which also means that we should be able to show that $\angle GMA$ is a right angle). Commented Aug 25, 2021 at 17:48
• What is a synthetic solution? @dodoturkoz Commented Aug 26, 2021 at 15:32
• @Buraian that's a good point :) I would call it geometric construction that leads to the answer without using trigonometry / equations. Here is something I found - math.stackexchange.com/questions/669037/… Commented Aug 26, 2021 at 15:47
• @Buraian I agree with MathLover. Though, many times the distinction between synthetic and analytic methods is not very clear (as we can derive trigonometric results using synthetic geometry). Commented Aug 26, 2021 at 16:12