sum of angles by 8 equilaterals (2012-13 Savin Competition 6-8 p10)

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sum of angles by 8 equilaterals (2012-13 Savin Competition 6-8 p10)

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#1 h Jun 16, 2021, 8:09 PM

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Eight equilateral triangles are located as shown in the figure. Find the sum of the two marked angles.

https://cdn.artofproblemsolving.com/attachments/c/f/34ef8c2cf3906a0c152028f676897026a3ffa9.png

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#2 h Jun 16, 2021, 8:09 PM

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#3 h Jun 16, 2021, 8:50 PM • 1 Y

Y by Mango247

Let the side of each triangle be $x$ . We will solve for the longest side for each of the two "corner" triangles:
$a^2=x^2+(2x)^2-2(x)(2x)\cos\left(120^\circ\right) \rightarrow a^2=5x^2-4x^2\left(-\frac{1}{2}\right) \rightarrow a^2=7x^2 \rightarrow a=x\sqrt{7}$ .
$b^2=x^2+(4x)^2-2(x)(4x)\cos\left(120^\circ\right) \rightarrow b^2=17x^2-8x^2\left(-\frac{1}{2}\right) \rightarrow b^2=21x^2 \rightarrow b=x\sqrt{21}$ .

Let the top-left angle be $\alpha$ and the bottom-right angle be $\beta$ . By the Law of Cosines again:
$(2x)^2=x^2+7x^2-2(x)\left(x\sqrt{7}\right)\cos{\alpha} \rightarrow 2x^2\sqrt{7} \cos{\alpha}=4x^2 \rightarrow \cos{\alpha}=\frac{2}{\sqrt{7}}$ .
$(4x)^2=x^2+21x^2-2(x)\left(x\sqrt{21}\right)\cos{\beta} \rightarrow 2x^2\sqrt{21} \cos{\beta}=6x^2 \rightarrow \cos{\beta}=\frac{3}{\sqrt{21}}$
This implies that $\tan{\alpha}=\frac{\sqrt{3}}{2}$ and $\tan{\beta}=\frac{2}{\sqrt{3}}$ . So $\tan{\alpha}\tan{\beta}=1 \rightarrow \tan{\alpha}=\cot{\beta} \rightarrow \tan{\alpha}=\tan\left(90-\beta\right)$ . We can set the inner parts equal to each other (we don't have to worry about coterminal angles and such since $\alpha$ and $\beta$ are acute), which gives $\alpha=90^\circ-\beta \rightarrow \alpha+\beta=\boxed{90^\circ}$ .

This post has been edited 1 time. Last edited by Bowser498, Jun 16, 2021, 8:59 PM
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#4 h Jun 17, 2021, 4:29 AM • 1 Y

Y by Mango247

Very nice!

Let the equilateral triangle have side length $1$ . Consider the leftmost marked angle. It is the angle opposite a side of length $2$ of a triangle with another side length $1$ . Let it be $\theta$ . The final side length is $\sqrt{\left(\frac{\sqrt3}{2}\right)^2+\left(\frac{5}{2}\right)^2}=\sqrt7$ . Now by Law of Sines we have $\frac{\sqrt7}{\sin 120^\circ}=\frac{2}{\sin \theta}$ . Then $\sin \theta = \frac{\sqrt3}{\sqrt7}$ . The angle is $\sin^{-1}{\sqrt{\frac{3}{7}}}$ . Similarly, letting the other angle be $\phi$ , we get $\frac{\sqrt{21}}{\sin 120^\circ}=\frac{4}{\sin \phi}$ . Then $\sin \phi = \frac{2}{\sqrt7}$ . The angle is $\sin^{-1}{\frac{2}{\sqrt7}}$ .

Adding these two gives $\boxed{90^\circ}$ .

This post has been edited 1 time. Last edited by OlympusHero, Jun 17, 2021, 4:30 AM

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#5 h Jun 17, 2021, 8:35 AM • 1 Y

Y by Mango247

My solution at https://artofproblemsolving.com/community/c4t48f4h2592875_sum_of_angles_by_8_equilaterals_201213_savin_competition_68_p10

Best regards,
sunken rock

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#6 h Jun 17, 2021, 2:27 PM • 1 Y

Y by BIBLEMATHS2015

Look at my attachment, I'll be using the letters in there.

Reflect the entire figure against line $AB$ . So we need to find $\angle{ZAF}$ .

Let the side of a triangle be $s$ . Hence by LoC on $\triangle{ZBF}$ , we have:
$ZF^2 = (4s)^2 + (2s)^2 - 2(2s)(4s) \cos 120^{\circ} = 16s^2 + 4s^2 + 8s^2 = 28s^2.$
Now by Pythagoras, let's say the altitude of $\triangle{ABK}$ is $AP$ so $AP = \frac{s\sqrt{3}}{2}$ and the altitude of $\triangle{ABC}$ is $AQ$ so $AQ = \frac{s\sqrt{3}}{2}$ . By Pythagoras we can see that $AF^2 = \frac{3s^2}{4} + \frac{25s^2}{4} = 7s^2$ and $AZ^2 = \frac{3s^2}{4} + \frac{81s^2}{4} = 21s^2$ .

Apply LoC on $\triangle{ZAF}$ to complete:
$ZF^2 = AF^2 + AZ^2 - 2AF \cdot AZ \cos \angle{ZAF} \iff 28s^2 = 7s^2 + 21s^2 - 14\sqrt{3} s^2 \cos \angle{ZAF} \iff 0 = 14\sqrt{3} s^2 \cos \angle{ZAF}$ which clearly means $\cos \angle{ZAF} = 0 \iff \angle{ZAF} = \boxed{90^{\circ}}$ (since both angles are acute, so their sum is less than $180^{\circ}$ ).

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#7 h Jun 17, 2021, 7:38 PM • 2 Y

Y by BIBLEMATHS2015, somebodyyouusedtoknow

With no calculations at all...

move and copy EB to CF on a triangular grid
then OE = OC = OF ensures triangle ECF is right angled at C (if a median equals half the opposite side, then this triangle is a right triangle = inscribed in a semi-circle with diameter EF)

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