by Donny Thurston

Today we are going to look at an exploration concerning angle bisectors of a triangle. For reference, the following problem is inspiring this exploration, although we will definitely move beyond the scope of the given problem.

7. The internal angle bisectors of triangle ABC are extended to meet the circumcircle at points L, M, and N, respectively. Find the angles of triangle LMN in terms of the angles A, B, and C. Does your result hold only for acute triangles?

Let's answer this question, but then explore what more we can learn about the angle bisectors in a triangle in this situation.

Let us consider the above situation, but use a figure found in a GSP sketch. You can follow along by making the simple constructions called for in the problem, or, at the end of this write-up, there will be a GSP file that includes the completed exploration.

In this figure, we will being with a circle and the triangle ABC. The triangle formed using the angle bisectors as described in the exercise will be called DEF. Throughout the write-up, we will usually refer to the angles using only the single letter (instead of saying "angle BAC" for instance), indicating that we mean the angle that is part of the whole triangle. We will only refer two angles with three letters when we are referring to smaller angles within triangles. The figure is as follows.

So, what do we know about angles D, E, and F? Well, given the properties of angles insribed within a circle, we can absolutely find angles D, E, and F in terms of angles A, B, and C.

Recall that the measure of an inscribed angle is equal to half of the measure of the intercepted arc. In addition, if two inscribed angles intercept the same arc, then those two angles must be congruent.

Also note that as an angle bisector splits an angle into two angles of equal measure, it must also split the intercepted arc into two arcs that are of equal measure. For instance, observe that angle ABE = 1/2 angle ABC. Therefore, since angle ABE intercepts minor arc AE, minor arc AE = 1/2 arc AEC.

Finally, if an angle were to intercept an arc that is the sum of two smaller arcs, then the measure of the angle must be equal to the sum of the measures of the two angles that intercept those smaller arcs. For example, angle D intercepts arc ECF which can be considered minor arc EC + minor arc FC. Since these arcs are equal to 1/2 B and 1/2 A respectively, we learn that the measure of angle D = 1/2 B + 1/2 A = 1/2 (B + A)

Using this reasoning, we can find each of the angles in triangle DEF in terms of ABC, answering the first part of the above exercise as follows:

E = 1/2 (C + A)

F = 1/2 (B + C)

D = 1/2 (B + A)

Or, if we prefer to use only a single angle from the original triangle in each equation, then I suggest this one:

E = 1/2 (C + A) = 1/2 (C + A + B) - 1/2(B) = -1/2(B) + 90

F = 1/2 (B + C) = 1/2 (C + A + B) - 1/2(A) = -1/2(A) + 90

D = 1/2 (B + A) = 1/2 (C + A + B) - 1/2(C) = -1/2(C) + 90

So, to answer the final part of the question, what about obtuse triangles? From looking at our equation or the figure, this does not seem to be a problem, but, just in case, observe the following figure:

In this instance, A is obtuse, yet the relationships still hold without question, as per our rules concerning inscribed angles.

This answers the initial question, but there appears to still be much that we can consider in this problem. Let us see if we can find a pattern.

Let us continue our exercise, and find another triangle, named HIG, following the same pattern as before, drawing angle bisectors through our second triangle and connecting their intersections with the circumscribed circle. Observe:

Observe that our previously discovered relationships hold between our second and third triangle as well, as seen here:

H = 1/2 (E + F) = -1/2 (G) + 90

I = 1/2 (D + F) = -1/2 (E) + 90

G = 1/2 (D + E) = -1/2 (F) + 90In fact, I think we can identify a function rule here. Let us consider A[n] any angle in the nth triangle constructed using this method of extending the angle bisectors. We can consider A[0] our initial triangle, ABC. In this case, it appears that:

A[n] = -1/2(A[n-1]) + 90 for all n is a subset of natural numbers > 0

Therefore, for every angle in a triangle created using this method, there is a corresponding angle in the previous triangle so that you can divide that angle by two and subtract it from 90 to get the A[n] angle.

Doing this would imply that:

A[n] = 1/4 A[n-2] + 45

Can we algebraically see that in action?

H = 1/2 (E + F) = 1/4 (A + B + 2C) = 1/4 C + 1/4 (180) = 1/4 C + 45

I = 1/2 (D + F) = 1/4 (A + 2B + C) = 1/4 B + 1/4 (180) = 1/4 B + 45

G = 1/2 (D + E) = 1/4 (2A + B + C) = 1/4 A + 1/4 (180) = 1/4 A + 45

So, does this tell us anything interesting about these triangles? Do the angles of the triangles follow some pattern, or do they just vary randomly? Let us construct a few more iterations, and see if we notice any unusual behavior.

Here we have two more triangles, JKL and MNO. Do we see any patterns?

Let me go ahead and suggest putting these new angles in terms of our original angles.

K = 1/2 (G + H) = 1/8 (3A + 2B + 3C) = 1/8 (A + C) + 1/8 (360) = 1/4 E + 45 = -1/8 B + 67.5

L = 1/2 (I + H) = 1/8 (2A + 3B + 3C) = 1/8 (B + C) + 1/8 (360) = 1/4 F + 45 = -1/8 A + 67.5

J = 1/2 (G + I) = 1/8 (3A + 3B + 2C) = 1/8 (A + B) + 1/8 (360) = 1/4 D + 45 = -1/8 C + 67.5

M = 1/2 (J + L) = 1/16 (5A + 6B + 5C) = 1/16 B + 1/16 (900) = 1/16 B + 56.25

N = 1/2 (J + K) = 1/16 (6A + 5B + 5C) = 1/16 A + 1/16 (900) = 1/16 A + 56.25

O = 1/2 (L + K) = 1/16 (5A + 5B + 6C) = 1/16 C + 1/16 (900) = 1/16 C + 56.25

What do we observe? Well, for starters, the "amount" of angles A, B, and C that are included in each of the succeeding angle calculations grows smaller and smaller in each interation of the trianlge. For instance, by interation MNO, only 1/16 of B (or any other angle of that generation) is included in the calculation. Everything else that makes up the angle is in the constant. In fact, that reveals another intersting feature can be seen. With each suceeding generation (as can be observed from the second until the current), the constant grows closer to 60 degrees. Does this mean something?

In fact, it does. It would follow that each new iteration of the triangle would more closely reflect anThe variable in each angle calculation that comes from the original trianlge (the A, B or C) represents the variance that that angle has. The more A, B, or C can vary, the more that angles calculated using A, B, or C can vary. However, as the values of A, B, or C are reduced through division, the less variance comes into the angles. The difference is made up by the consants growing closer and closer to 60 degrees.equilateral triangle.Can we conceptualize this with examples? Sure! Consider the first generation. In that instance, angle A, B, or C can vary up to 120 degrees (but NOT including 120) from 60 degrees. Angle A, for example, could become very close to 180 degrees, although it could not become exactly 180 degrees. If that were the case, then angle F (see how F is dependent on A above) could vary by up to

half of thatsince A is divided by 2. Angle F could vary by up to (but not including) 60 degrees from 60.Consequently, since A (in our example here) is divided by four in the 3rd iteration, HIG, then angle G can vary by up to 120/4 or 30 degrees. After 5 generations, (our MNO), the greatest an angle can vary is 120/16 = 7.5 degrees.

As each generation passes, the amount that an angle, at most, can vary from 60 degrees is cut in half. As a result, we can construct a system that looks like this:

V[n] = amount an angle can vary from 60 degrees at generation n.

V[1] = 120

V[n] = V[n-1]/2

We see from that:

6th iteration will be within 3.75

7th iteration will be within 1.875

8th iteration will be within .9375

9th iteration will be within .46875So, how long must we wait until we get a triangle that is equilateral? Strictly speaking, it would never happen (unless the initial triangle was equilateral), but effectively, it all depends on how close you need to get. Within one degree? 8th iteration. Within a half of a degree? 9th. After 10 iterations you are within a quarter of a degree.

Find the GSP sketch used in this write-up here. Measurements are included and you can see it all at work!