Name:
Date:
Honors Math 3 investigation
extension to Chapter 9A day 2, page 1
Solution to the Highway Optimization Problem
Constructing the equiangular point inside a triangle
Introduction: In our investigation of the Highway Optimization Problem, we discovered
experimentally that for many triangles, the optimal location for the highway junction was the
equiangular point: the point Q inside ABC such that the three angles AQB, BQC, and
CQA are all equal (specifically, all 120º, since the sum of these angles is 360º). This raises the
question of whether there is a construction of the equiangular point. Besides allowing us to create
the point exactly on the computer, a construction would also guarantee that an equiangular point
must exist inside the triangle.
Directions: Perform the construction steps listed below using GeoGebra. After certain steps,
there are proof questions to answer in writing on separate paper. Taken together, these questions
will form a proof that the constructed point is the equiangular point.
Important warning: We are using GeoGebra to build the diagram, but we cannot use computer
measurements as evidence in our proofs. All proof steps need to be
justified using postulates or theorems that we have studied.
Step I. Begin with ABC. (For now, try an acute triangle.)
Step II. Construct an equilateral triangle GAB as shown.
Details: Draw a circle centered at A passing through B, and a circle
centered at B passing through A. Locate G at an intersection of the
two circles, then hide the circles (by control-clicking or
right-clicking, then turning off “Show Object”).
diagram after Step II
Step III. Construct a circle passing through the vertices of GAB.
Question 1. Suppose J is any point along minor arc AB. Prove that
mAJB = 120º. [Then hide J before going on to Step IV.]
Step IV. Repeat Steps II and III along side AC to create an
equilateral triangle HAC, and a circle passing through the vertices
of HAC.
diagram after Step III
Question 2. Suppose K is any point along minor arc AC.
Prove mAKC = 120º. [Hide K before going on to Step V.]
Step V. There should now be two circles in the diagram.
These circles intersect at A and at a second point Q. Below
you will prove that Q is the desired equiangular point.
Question 3. Draw segments QA, QB, and QC.
a. Prove that mAQB = 120º. Hint: See Question 1.
b. Prove that mAQC = 120º. Hint: See Question 2.
c. Prove that mBQC = 120º. Hint: Use parts a and b,
and the 360º angle sum.
Thus you have shown that Q is the equiangular point.
diagram after Step V
Name:
Date:
Honors Math 3 investigation
extension to Chapter 9A day 2, page 2
Question 4. Can ABC have any other equiangular points besides Q? Justify your answer.
Question 5. This equiangular point construction applies to acute triangles, and some other
triangles, but not all triangles. By computer experimentation and/or by scrutinizing the
construction, determine exactly what triangles ABC would allow this construction to work.
Question 6. For triangles where this construction does not work, how and why does the
construction fail? In these cases, does an equiangular point still exist, or does the triangle not
have an equiangular point at all?
Homework: Proof that the equiangular point is optimal
The purpose of this homework is to verify that in triangles where an equiangular point exists, the
solution to the Highway Optimization Problem is that building highways to the equiangular point
produces the minimum total road length.
An outline of a proof is given below. Write out the proof, filling in all the details. Draw the
diagram for the proof either using GeoGebra (preferred) or by hand.
Given: Any ABC with an equiangular point Q.
To prove: Q is “better” than any other point R inside ABC, in the sense that
QA + QB + QC < RA + RB + RC.
First part of proof Add a triangle XYZ into the diagram, and show that it is equilateral.
Construct the line through A perpendicular to QA, the line through B perpendicular to QB, and
the line through C perpendicular to QC. The three lines form a new triangle; call it XYZ. Prove
that XYZ is equilateral. Hint: Show that the angles of XYZ all must equal 60°.
Second part of proof Show that Q is “better” than any other point R.
Choose any point R (not coinciding with Q) in the interior of ABC. The goal is to prove that
QA + QB + QC < RA + RB + RC. Here is an outline of the proof:





Construct the perpendiculars from R to the three sides of XYZ. Name the three points of
perpendicularity as follows: let D be the point lying on the same side as A, E the point
lying on the same side as B, and F the point lying on the same side as C.
Use Rich’s Theorem (from day 1, page 3) to show that QA + QB + QC = RD + RE + RF.
Explain why RD  RA, RE  RB, and RF  RC.
Combine the two previous steps to get that QA + QB + QC  RA + RB + RC.
Argue that QA + QB + QC cannot equal RA + RB + RC.