The Math Forum

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

Incircles Tangent to a Common Line

Date: 03/23/2001 at 00:20:23
From: Alice Klein
Subject: Circle Geometry

In triangle ABC, the incircle touches side AB at M. T is an arbitrary 
point on BC. Show that the incircles of triangles BMT, AMT and ATC are 
all tangent to a common line.

This problem proved really difficult to me, I could not think of a 
good way to approach it. Can you help me?

Thanks heaps.

Date: 03/26/2001 at 03:33:38
From: Doctor Floor
Subject: Re: Circle Geometry

Hi, Alice,

This indeed is a difficult problem.

We will repeatedly use the fact that the distance from a point of 
tangency of an incircle to a vertex is half the perimeter of the 
triangle minus the side opposite to the vertex. So for instance, we 
write AM = s-a where s means (a+b+c)/2 (a is the side opposite to A).

See for a short explanation the following message from the Dr. Math 

   Geometric Proof of Heron's Formula   

(It is item 1. in the proof).

Now let's consider your figure:


In this figure I have drawn a tangent to all three circles UWY, but we 
are not sure that it really is a line. It could be that there are 
three segments not contained in one line. For ease of naming I will 
restrict myself to considering the three tangents UW, WY, and UY, 
while in fact the points of tangency to the circle (D) (with center D) 
could be different for tangents UW and UY.

Now I will show that:

  1. the added lengths of UW and WY are equal to the length of UY;

  2. there is only one choice of M, for fixed T, that satisfies 1.

  3. This proves the theorem, because:

      * when M is very close to A, then W is closer to A than to UY
        (note that W is then in fact two points),

      * when M is very close to B then UY is closer to A than to W,

So there must have been a choice for M that W lies on UY. For that 
choice the lengths UW + WY and UY must be equal.

Proof of 1:
We use the s-a alike statements:

   a.    GT = 1/2(AT + TC - b)
         IT = 1/2(MT + TB - MB)
      so GI = 1/2(AT + MT - BM + a - b)
         (since TC + TB = a)

   b.    TK = 1/2(MT + TB - BM)
         MJ = 1/2(AM + MT - AT)
      so JK = MT - TK - MJ = 1/2(BM + AT - AM - TB)

   c.    AL = 1/2(AM + AT - MT)
         TH = 1/2(AT + TC - b)
      so HL = AT - AL - TH = 1/2(b + MT - AM - TC)

   d. Thus HL + JK = 1/2(BM - 2AM + AT + MT - a + b)

   e. When we substitute AM = s-a and BM = s-b in the results of a. 
      and d., both turn out to be equal to 1/2(AT + MT + a - s)

   f. A pair of external tangents to a circle is congruent, as well as 
      a pair of internal tangents. So UY = GI, UW = LH, and WY = JK. 
      And thus 1 is proven.

Proof of 2:

   a. We note that if M moves, then AM increases with some number t, 
      while BM decreases with the same t and MT increases with some 
      number u (where t and u might be negative);

   b. Applying this to the formula for GI in 1a, we find that GI 
      increases by u + t;

   c. In the same way, using d we see that HL + JK increases with 

   d. Essentially the difference between UW + WU and UY changes by 4t 
      when M moves through a distance of t. So there can be only one 
      point where the sums are equal.

That should be the proof.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum   

Date: 03/26/2001 at 04:45:17
From: George Zhong
Subject: Re: Circle Geometry

Hmmm, yes, I see what you mean, Dr Floor.

But could you explain how you proved part 1:

  1. That the added lengths of UW and WY are equal to the length 
     of UY.

in greater detail?

You lost me halfway through. Sorry to take up your time with these 

Best regards.

Date: 03/26/2001 at 07:51:27
From: Doctor Floor
Subject: Re: Circle Geometry

Hi, George,

Thanks for your response.

What I did (and for that I pointed you to the Dr. Math archives) was 
use the following theorem:

       v/ \v
       Z   Y
     t/     \u
       t   u

The points X, Y, and Z are the points where the incircle meets the 
sides of ABC.

A pair of tangents from one point to a circle are congruent, so we 
have PX = PZ = t, QX = QY = u, and RZ = RY = v. So t+u+v is half the 
perimeter, and thus, u is half the perimeter minus PR.

So u = 1/2(PQ+QR+PR)-PR = 1/2(PQ+QR-PR).

Then, perhaps for 1d you didn't realize that (just as in 1a) I again 
used that TC + TB = a.

I hope this clears it up. If not, just write back.

Best regards,
- Doctor Floor, The Math Forum   
Associated Topics:
High School Conic Sections/Circles
High School Geometry
High School Triangles and Other Polygons

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.