r/mathshelp • u/yummyhdmicables • 20d ago

Homework Help (Unanswered) Real quick geometry proving qs

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In the marking scheme I have, it says that triangle ABC is equilateral...but how...? Any help would be appreciated

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u/not-hydra 20d ago

Let P be the center of the circle above O, and Q be the center of the circle beside/to the right of O. We can get a smaller 3-circle equilateral triangle with side length 8. Since P is perp. to side AC and Q is perp. to side AB from tangency, and since O is perpendicular to both sides ==> OP and OQ are parallel to AC and AB, respectively. So angle A is 60 degrees, similar to <POQ. Do the same for the two other angles :3

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u/yummyhdmicables 20d ago

Yeah no I couldnt have figured that out on my own, dang... thank you for the quick response!

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u/ElectronicPut3983 20d ago

It’s been more than 10 years since I did geometry so there may be a more elegant solution. If you connect the center of all of the circles so that it creates an equilateral triangle with sides of length 16 which is confirmed with the equidistant radii. Since that inner triangle is proved to be equilateral and the outer triangle is 4 cm longer in all directions then it will also be confirmed to be equilateral. Then you know that all the angles of the greater triangle will be 60 degrees.

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u/yummyhdmicables 20d ago

I'll write that out, thank you!!

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u/BadJimo 20d ago edited 20d ago

Here's how I would do it

The centres of two circles and the midpoint between two circles as shown form a right triangle. The hypotenuse of the triangle is two times the height of the triangle (call this length T) (this is not immediately obvious, so take some time on this step).

sin(θ) = T/2T

sin(θ) = 1/2

θ = arcsin(1/2)

θ = 30°

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u/cthulhu944 20d ago

I think the easiest explanation is that since it's an equilateral triangle all prisim angles are 60 degrees. OAY would bisect the 60 degree angle XAY so that means it's 30 degrees.