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Chapter 2: Problem 5

The circumradius of a triangle is at least twice the inradius.

Short Answer

Expert verified
The circumradius \(R\) of a triangle is always at least twice the inradius \(r\). This statement is proved by connecting the values of R and r to the triangle's area, and using the inequality that the semiperimeter \(s\) is always greater or equal to twice the circumradius. After canceling out common factors, we're left with the inequality \(R/r 鈮� 2\), confirming the initial statement.

Step by step solution

01

Rewrite the equation for area

Let's write down the area of a triangle using both the inradius and the circumradius: area = rs = abc/4R.
02

Isolate R to one side

Isolating R we have R = abc/4rs.
03

Insert inequality s 鈮� 2R

Now we know that s, the semi-perimeter of the triangle, is greater than or equal to 2R. So we replace s with 2R on the left side: abc/4r 鈮� 2abc/4R = 2Rs.
04

Cancel out common factors

Now we cancel out the common factors on both sides, so we're left with: R/r 鈮� 2.
05

Interpret the Result

This means that the circumradius R must be at least twice the inradius r.

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