I apologize for the name, I just do not know what to call the topic. The thing here is what ..
Imagine that you have an equilateral-right angle and a plane on which this angle lies on one side. Now find the imaginary center and turn this angle so that its second side takes the place of the first. I understand that it is not clear why the illustration below.

enter image description here

When I turn the angle 90 degrees around point O , the segment CA will take the place of segment AB .

Now imagine that the angle is 45 degrees.

enter image description here

How can I find the length of the OE radius so that putting the perpendicular of equal length from the segment AB , I found a point, turning the angle around which I placed the segment CA in the position AB at 45 degrees?

  • I want to add that the second drawing was painted on the eye and the side at an angle of 45 degrees is not true. - vas
  • draw the starting and ending position of the angle. I do not understand from your explanation of the problem: ( - splash58
  • and where does the perpendicular plane, if everything happens in one plane? - splash58
  • @soon And Qwertiy did not write about the same thing? - splash58
  • @ splash58, yes, that's right. For some reason, SO did not show a plate with new answers - awesoon

2 answers 2

If I understood the condition correctly, then the point O should be placed at the intersection of the bisector of the angle BAC and the median perpendicular to the segment AB.

In general, an isosceles triangle AOB with base AB, the angle OAB is equal to half the angle BAC. It is necessary to find the height. To do this, there are several ways of varying complexity.

  • Can I have a formula? - vas
  • @shatal, oops .. the formula was wrong. - Qwertiy ♦
  • after reading the answer below, I wanted to say thank you for the detailed explanation. - vas

So it's very simple, I think. Denote by x half of segment AB , it will be half АС , radius R should be such that the circle touched segment AB, and segment AC, dividing them in half at the points of tangency (proof of this fact is discussed in the 9th grade on the geometry that the tangent issued from one point have equal length until the circle touches the circle). Next we get the equation x/R = tg Pi/8 . Hence R = x / (tg Pi/8) . The value of x I understand, you know. The tangent of Pi/8 is equal to the root of the particular two differences: the first difference is the root of two minus one, and the second root of two plus one.