Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given a circle whose center (P) is at center of a square, and using pencil, compass, and straight-edge, how can I create a right triangle whose hypotenuse is side of square and whose side is tangent to the circle?

I can get side to be tangent to the circle and I get the sides to be right angles but I'm not able to get both right angle and tangent. This is not homework.

enter image description here

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Let $ABCD$ be the given square, $P$ the centre of the circle. The circle with diameter $AP$ intersects the circle in a point $Q$. The line $AQ$ intersects the circle with diameter $AB$ in a point $R$. Then $ABR$ is a rectangular triangle with hypothenuse $AB$ and with $AR$ tangent to the given circle.

share|improve this answer
    
Hagen: That works for me...sweet and simple. –  zundarz Feb 8 '13 at 23:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.