# Geometry problem

**URL:** <https://boards.straightdope.com/t/geometry-problem/483813>\
**Category:** Factual Questions\
**Created:** [January 31, 2009, 1:38am UTC](https://boards.straightdope.com/t/geometry-problem/483813 "2009-01-31T01:38:28Z")\
**Posts on this page:** 6\
**Page:** 1

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**Author:** ![lissener](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lissener/32/6139_2.png) [@lissener](https://boards.straightdope.com/u/lissener)\
**Post date:** [January 31, 2009, 1:38am UTC](https://boards.straightdope.com/t/geometry-problem/483813/1 "2009-01-31T01:38:28Z")

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I used to know how to do this. In fact, I specifically remember doing it when I was designing a triangular knit hat. But I can’t for the life of me remember how anymore.

Can someone remind me how to find the [dimension](http://img204.imageshack.us/img204/2526/triangleyd4.jpg) of the outside triangle based only on the radius of the enclosed circle?

Thanks.

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**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\
**Post date:** [January 31, 2009, 1:49am UTC](https://boards.straightdope.com/t/geometry-problem/483813/2 "2009-01-31T01:49:53Z")

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Draw a radius going to the center of one of the triangle’s sides, and then draw a line from the center of the circle to one vertex of the triangle. That gives you a right triangle where one leg is the radius of the circle, and the other leg is half the length of the big triangle’s side. The pointy angle of that right triangle is 30 degrees, and from there it’s all trigonometry.

Jumping to the end of the problem, we find that the side of the big triangle is 2\*sqrt(3) times the radius of the circle.

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**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\
**Post date:** [January 31, 2009, 1:52am UTC](https://boards.straightdope.com/t/geometry-problem/483813/3 "2009-01-31T01:52:37Z")

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I took time to add “Assuming the enclosing triangle is equilateral” to my answer, and was scooped by Chronos.

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**Author:** ![RickJay](https://avatars.discourse-cdn.com/v4/letter/r/bb73d2/32.png) [@RickJay](https://boards.straightdope.com/u/RickJay)\
**Post date:** [January 31, 2009, 1:53am UTC](https://boards.straightdope.com/t/geometry-problem/483813/4 "2009-01-31T01:53:11Z")

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> [@lissener](#):
>
> I used to know how to do this. In fact, I specifically remember doing it when I was designing a triangular knit hat. But I can’t for the life of me remember how anymore.
> 
> Can someone remind me how to find the [dimension](http://img204.imageshack.us/img204/2526/triangleyd4.jpg) of the outside triangle based only on the radius of the enclosed circle?
> 
> Thanks.

Assuming your diagram describes an equilateral triangle, the radius of an incircle is

radius = ? \* (root 3)/6

so

4= ? \* .288675  
? = 13.86

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**Author:** ![lissener](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lissener/32/6139_2.png) [@lissener](https://boards.straightdope.com/u/lissener)\
**Post date:** [January 31, 2009, 1:54am UTC](https://boards.straightdope.com/t/geometry-problem/483813/5 "2009-01-31T01:54:58Z")

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I think it’s 3.4641"

No?

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**Author:** ![lissener](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lissener/32/6139_2.png) [@lissener](https://boards.straightdope.com/u/lissener)\
**Post date:** [January 31, 2009, 1:56am UTC](https://boards.straightdope.com/t/geometry-problem/483813/6 "2009-01-31T01:56:14Z")

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oh wait, times 4

13.8564

So yeah, RJ is right. And I remembered how to do it!