Intermath
Title
Optimizing Triangles Using Radii
Problem Statement
A triangle is inside a circle. The circle's center is one vertex of the triangle, and two
of the circle's radii represent the legs of the triangle. What central angle is necessary
to produce a triangle with maximum area? With minimum area?
Problem setup
I want to find the central angle of a triangle (that is inscribed inside of a circle) that will
produce both the maximum and minimum area for a triangle. I will use the formula for
finding the area of a triangle, A = ½ bh.
Plans to Solve/Investigate the Problem
I believe that I am going to have to play around with this problem. I will construct a
circle. I will need to use the same circle each time so that each angle I create is accurate.
I will start with a smaller central angles and reaching a larger central angle: 20°, 40°, 60°,
80°, 90°, 100°, 120°, and 140°, 160°, and 175°. I do not know a pattern to approach this
situation. Experimenting with each of these central angle measurements, I feel as If I will
find either a pattern and/or the answer to my problem.
Prediction: I predict that the central angle that will produce the largest maximum area is
175° because it is the largest central angle I can create and measure. I predict the
smallest minimum area that is produced will come from the smallest central angle
possible.
Investigation/Exploration of the Problem:
I constructed a circle. In order to keep this circle the same size throughout my
investigation, I copied the circle several times.
For the central angles of 120° and 60°, I created a circle by center + radius after
connecting the center vertex and a random vertex on the circle. For these two, I created
circles around center B and center E, using AB and EB . Before I measure the angles, I
created ABC and EFB . I created B around G , creating GB and then constructing a
circle by center + radius. I then rotated GB 90° around vertex G, connected the vertices,
and constructed GBB ' .
To measure the areas of each triangle, I create a midpoint on the base of each triangle,
construct a line segment, and measure it. This gives the height of the triangle. I measure
the base of the triangle to give me the base. For example, in the first circle created, with
central angle 120°, I created midpoint D, then created BD , measured it and got my height.
I measured AC and measured it for my base. I then used the area formula for a circle to
get the area of ABC .
mABC = 120.00
m BD = 0.77 cm
mFEB = 60.00
mBGB' = 90.00
m EK = 1.33 cm
m CA = 2.66 cm
0.5m CAm BD = 1.02 cm 2
m B'B = 2.17 cm
0.5m FBm EK = 1.02
cm 2
B'
C
B
E
D
A
m HM = 1.51 cm
G
K
F
mBHB' = 20.00
B
L
mBOB' = 80.00
mBIB' = 40.00
m NI = 1.44 cm
0.5m B'Bm NI = 0.76
I
B'
M
B
m QO = 1.18 cm
m B'B = 1.97 cm
m B'B = 1.05 cm
0.5m B'Bm HM = 0.40 cm 2
H
0.5m B'Bm GL = 1.18 cm 2
B
m B'B = 0.53 cm
0.5m B'Bm QO = 1.16 cm 2
cm 2
B'
O
B'
Q
N
B
B
mBAB' = 100.00
mBCB' = 140.00
m AB = 0.99 cm
0.5m B'Bm AB = 1.16
B'
0.5m B'Bm CD = 0.76 cm 2
cm 2
B
C
B
mBEB' = 160.00
m CD = 0.52 cm
m B'B = 2.88 cm
m B'B = 2.35 cm
B'
A
m GL = 1.08 cm
m FB = 1.53 cm
D
B
m EF = 0.27 cm
mBGB' = 175.00
m B'B = 3.07 cm
m B'B = 3.02 cm
B'
0.5m B'Bm EF = 0.40 cm 2
F
E
m GH = 0.07 cm
B'
0.5m B'Bm GH = 0.10 cm 2
H
G
B
B
Looking at these pictures, I see that the circle that has the triangle with the largest central
angle is the 90° central angle. Therefore, my prediction is way off.
Taking a closer look at the triangle with the largest central angle:
m AB = 1.53 cm
m CA = 1.08 cm
m AC = 1.53 cm
m CB = 2.17 cm
m CB = 2.17 cm
0.5m CBm CA = 1.18 cm 2
mBAC = 90.00
A 90° angle, when reflected across it self,
forms a square. Therefore, the largest area is
formed by a right angle isosceles triangle.
C
A
C
B
Extension of the problem:
I want to know what happens when all vertices are placed on the rim of the circle.
I created Z , and then created AC so that vertex A moves concurrently with vertex B. I
constructed vertex B randomly on Z and connected segments to form ABC . When I
measure the triangle, I discovered that no matter where vertex B lied on Z , the measure
of ABC remained 90°. I have shown examples below to show that no matter where
vertex B lies, the triangle remains an isosceles triangle with a 90° central angle.
mABC = 90.00
mABC = 90.00
mABC = 90.00
A
B
A
Z
C
A
B
C
Z
ABC = 1.63
cm 2
Area
ABC = 0.46
cm 2
Area
A
B
Z
C
Area
mABC = 90.00
Z
C
B
ABC = 1.76
cm 2
Area
ABC = 2.08 cm 2
A circle measures 360°. Because Z is divided in half
mBZG = 135.35
mBZA = 44.65
mBZG+mBZA = 180.00
mABC2 = 180.00
>
by AG , each half of Z measures 180°. By definition, the
sum of the measures of the interior angles in a triangle equal
180°. A right angle measures 90°. Notice that ABG is
divided in half by BZ . Adding the interior angles shows
that the triangle equals 180°.
mABC = 90.00
G
C
Z
A
Area
Author and Contact:
Lauren Mofield, Middle Grades Cohort
Darlnlulu4@yahoo.com
B
ABC = 2.35 cm 2