Assignment # 6
The Locus of a Vertex
By : Elizabeth Gore

The problem we will
be exploring in this investigation is:
HOW CAN WE FIND THE
LOCOS OF A POINT WHEN GIVEN TWO VERTICES AND THE ANGLE
AT THE THIRD VERTEX
OF A TRIANGLE ABC ??

First we will begin
by constructing the triangle ABC with the circumcenter and circumcircle.
I have also gone ahead
and connected the vertices to the circumcenter, C1.

Using our knowledge
of triangles and circumcenters, we know that c1 is equidistance
from each of the vertices.
So we have d(Ac1) = d(Bc1) = d(Cc1).
It is easy to infer
that triangle ABci is an isosceles triangle, because we know that
d(Ac1) = d(Bc1)
The Isosceles Triangle
Theorum tells us that since d(Ac1) = d(Bc1),
then the measurement
of angle (c1AB) must equal the measurement of (c1BA).
If we think about it
carefully, we can call these angles the same angle when speaking
of measurements.
So, we can conclude
that 2m(c1AB) + m (Ac1B) = 180 degrees. (Triangle sum Theorum)
Similarly, we can show
that 2m(CBc1) + m(Cc1B) = 180, and
2m(CAc1) + m(Cc1A)
= 180.
The actual angle measurements
are listed below the sketch.

Just as we saw with
triangle Ac1B, we can apply the Triangle Sum Theorum to triangle
ACB.
m (ACB) + m(CBA) +
m(BAC) = 180 degrees

Since we know that
angle measure is additive, we can conclude the following:
m(CAB) = m (CAc1) +
m(c1AB)
m(ABC) = m(ABc1) +
m(CBc1)
m(BCA) = m(BCc1) +
m(c1CA)
Now it is fitting for
us to say that
2m(c1AB) + 2m(c1BC)
+ 2m(c1CA) = 180 degrees
The actual angle measurements
are listed below. I will leave the computation for you!
We can see from the above picture that
m(Cc1A) + m(Ac1B) +
m(Bc1C) = 360 degrees
With some rearranging
we get:
m(Ac1B) = 360 - m(Cc1A)
- m(Bc1C)
m(AciB) = 360 - (180
- 2m(c1CA)) - (180 - 2m (BCc1))
m(AciB) = 2(m(c1CA)
+ m(BCc1))
From our investigation,
we saw that
m(BCA) = m(BCc1) +
m(c1CA)
So through manipulation
of the different angle equalities, we get that
m(BCA) = m(c1BC) +
m(c1CA)
Therefore:
m(Ac2B) = 2m(ACB)

Keeping in mind the
knowledge we have built up to, it follows that for any point H
on the given circle,
which is on the same
side of the chord AB as point C, with H not C,
m(Ac1B) = 2m(AHB).
Click HERE
for a gsp sketch to see that the measurement of angle AHB will
stay the same as long as you stay on
the "C" side of chord AB.

The length of segment
AB is constant, thus m(Ac1B) is constant.
Now we can make the
conclusion that
the locus of point
C is the major arc of angle (Ac1B).
If you are thirsty
for more, try and figure out if anything special happens when
point H dips below the chord.
If not, Click HERE to return to my homepage!