The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Triangle Construction

Date: 03/11/2002 at 03:29:26
From: Louise Hjarnaa
Subject: Construction of a triangle

Let ABC be a triangle with sides a, b, c. Let r be the radius of the 
incircle and R the radius of the circumcircle. How can I construct the 
triangle, in a classical way using only ruler and compass, when I know 
a, R, r ?

Date: 03/11/2002 at 05:07:31
From: Doctor Floor
Subject: Re: Construction of a triangle

Hi, Louise,

Thanks for your question.

Start with segment BC of length a. We know that the circumcenter O 
must lie on the perpendicular bisector p of BC: construct p. We also 
know that the distance from B to O must be equal to the given length 
R, so draw the circle with radius R and center B, intersect it with p 
and find O.

Now that we have O, we can draw the circumcircle. Now we choose A to 
be on either side of AB (both sides may yield possible triangles, so 
you must repeat the construction for the other side). On that side 
construct line d parallel to BC at a distance of the given r. The 
incenter I must be on this line d. 

Let M and M' be the points where p meets the circumcircle: M at the 
same side of BC as A, M' at the opposite side. The angle bisector from 
A must bisect the arc BC of the circumcircle opposite to A, so the 
angle bisector passes through M'.

Since all possible A on this side of BC yield the same angle <BAC, we 
see that the angle <BIC is fixed as well, since 

        <BIC=180-(<B+<C)/2= 180-(180-<C)/2. 

But that means that the possible I pass through an arc. This arc 
contains B and C, since these are limit points of I when A approaches 
B or C. Of course the arc also contains the incenter I' of triangle 
BMC, which we can construct easily.

So I must lie on r as well as on the circle through B, C and I'. The 
two intersections give the possible I. Intersecting M'I with the 
circumcircle gives the possible A, as desired.

That gives A on one side of BC. Of course you will have to repeat the 
construction for the other side, which may yield more solutions.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum   

Date: 03/12/2002 at 03:30:34
From: Louise Hjarnaa
Subject: Construction of a triangle

Dear Doctor Floor,

Thank you for your nice and very fast answer to my problem. 

It seems that the center of the circumcircle for BI'C is always  
point M'. Why?

Best regards and thanks in advance,

Date: 03/12/2002 at 10:15:14
From: Doctor Floor
Subject: Re: Construction of a triangle

Hi, Louise,

Let's see. Angle BIC = 180 - (<B+<C)/2 = 180 - (180-<A)/2= 90 + <A/2. 

Since M' is on the circumcircle of ABC opposite to A, <BM'C= 180 - <A. 
Now let's produce segment IM' to meet again the circumcircle of BIC 
in Q. Then of course <BQC = 180 - <BIC because they are opposite 
angles in a cyclic quadrilateral. But then

  <BQC = 180 - (90 + A/2) = 90 - A/2.

The central angle on BC in the circle (BICQ) must now be 
2*<BQC = 180 - <A. From this we see that the circumcenter of ABC must 
be on the arc BM'C. Also it must be on the perpendicular bisector of 
BC (triangle BM'C is isosceles). The point of intersection of these is 
M', and thus M' is indeed the center of BIC.

Best regards,
- Doctor Floor, The Math Forum   
Associated Topics:
College Constructions
High School Constructions

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.