Finding the Trisection of an Angle

ѳ

r

r

By: Rebekah Rudd Brian Sierra Priscilla Ferrat

Draw a line from point Q, the point where line OA intersects the circle, to a point X outside the circle. Point P is the intersection of the line and the circle. Segment XP is equal to r.

ѳr

rO

AQ

XP

r r

ѳr

rO

AQ

X r

We can see that triangle XPO is isosceles since segments XP and PO are both of length r, and also that triangle QOP is isosceles since its segments PO and QO are also of length r.

αr

P

Now we can label angles PXO and POX as equal to α and angles QPO and OQP as β.

β

ѳr

rO

AQ

X rα

rP β

Since the interior angles of a triangle add up to 180, angle XPO=180-2α.

As we can see, β is the supplementary angle to XPO. Therefore we can say that β+XPO=180.

Now we continue with only the angles and our equalities.

ѳ

O

αβ

XPO=180-2α.

β+XPO=180.

First we find β in terms of α.

β+(180-2α)=180

β-2α=0

β=2α

P

Q

X

A

ѳ

O

αβ

XPO=180-2α.

β+XPO=180.

Next we find ϴ using the fact that angle (QOP+α) is the supplementary angle of ϴ.

β=2α

O

Q

P

Angle QOP=180-2β.

Therefore α+(180-2β)+ϴ=180

α-2β+ϴ=0α-4α+ϴ=0

-3α+ϴ=0

ϴ=3α X

A

ѳ

O

αβ

O

SUCCESSFUL TRISECTION

ϴ=3α