Geometry

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TANTON’S TAKE ON …

CIRCLE THEOREMS

APRIL 2014

Here’s a cool activity/puzzle:

Take a piece of paper and push it up

between two nails in the wall. Mark

where the 90 o corner of the paper lies.

And do this again, pushing the paper up

between the nails at a different angle.

And then again, another fifty times.

What curve do the corners of the paper

seem to trace?

Suppose we conducted the same activity

but with a piece of paper cut to have a

corner angle different from 90 o . What

shape curve is being traced by this

corner?

Really do try these activities. (Draw two

dots on a white board rather than using

nails!)

Can you prove any claims you are

tempted to make?

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SOME PAPER-PUSHING THOUGHTS

In trying the first paper-pushing puzzle

with a 90 o corner one is very tempted

to say that the curve traced is a

semicircle. In fact, one would even say

that the center of that semi-circle is the

midpoint of the line segment connecting

the two nails.

One way to see that this is correct is to

draw the right triangle formed by that

line segment and the paper, and also a

rotated copy of that triangle

underneath it.

The two right triangles together make a

rectangle. (Why?)

We learn in geometry that the diagonals

of any rectangle are congruent and

bisect one another.

It follows that corner of the paper is

sure then to lie on the semicircle with

diameter defined by the nails.

Question: Is the answer to the second

puzzle also an arc of a circle?

These paper-pushing puzzles can

motivate a study of circles and

theorems about them.

SOME CIRCLE THEOREMS

What is the shortest route from a

point to line? (Assume the point in

question is not on the line.)

Most people would say that the straight

path that meets the line at a right angle

provides the shortest route.

How can we prove this claim?

One approach is to argue that no other

straight route could be shorter. For

example, look at the two routes of

lengths a and b offered in this

diagram:

We see a right triangle and so, by the

Pythagorean Theorem we can say:

( ) 22 2

b a something else= + .

Thus 2 b is larger than

2 a . And since we

are dealing with positive quantities, it

follows that b is larger than a . The blue

path is indeed longer than the

perpendicular red path.

This same argument proves that any

straight path to the line different from

the red path is sure to be longer in

length.

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A Philosophical Question: So … has this

proved the claim? All we have actually

shown is: Any path that is not the red

path is not the shortest path.

What do you think of this next theorem

and its “proof.”

Theorem:

1 is the largest counting number.

Proof: We’ll show that any counting

number N that is not 1 cannot be the

largest counting number. (This then

leaves 1 itself as the only available

option as the largest counting number.)

If N is a counting number different

from one, then 1N > . Multiply through

by N to get 2

N N> . Thus 2

N is a

counting number bigger than N ,

proving that N isn’t the largest

counting number.

OUR FIRST CIRCLE THEOREM

We’re all set for our first circle result.

Consider a circle, a tangent line to the

circle, and the radius that meets the

tangent line at its point of contact.

It is clear that any other path from the

center of the circle to point on the

tangent line is longer than the radius of

the circle. Thus the radius segment to

the point of contact is the shortest path

from the center of the circle to the

tangent. By the opening exercise, it

must meet the tangent line at a right

angle. We have:

Radius/Tangent Theorem:

Consider now two tangent lines to a

circle, or at least two tangent line

segments from a common point outside

the circle. The radii to the points of

contact meet these tangent segments at

right angles.

Exercise: Prove that the four corners of

the quadrilateral we see are con-cyclic,

that is, they all sit on a circle.

It seems compelling to ask: Does length

a equal length b ?

If we draw the line shown, labeling its

length c , say, then we see that the

answer is YES!

2 2

2 2

a c r

b c r

= −

= −

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We have:

Two Tangents Theorem:

EXERCISE: What the value of w?

(Assume we have tangent line

segments, tangent at the common

points of contact of the circles.)

Exercise: In drawing this hexagon that

circumscribes a circle (each side of the

figure just touches the circle) did I use

more red ink than blue ink, or more blue

ink than red?

To me, the most astounding circle

theorem of all is the following:

THE OPERA HOUSE THEOREM: All

peripheral angles subtended from the

same arc have the same measure.

Question: I used old-fashioned language

in the statement of the theorem. Can

you and your students make good

educated guesses as to what

“peripheral angles” are and what

“subtend” could mean?

Request: Can we teach students the art

of deducing the meaning of jargon? (I’d

vehemently object to “subtend” and

“peripheral” being reduced to vocab

words for a quiz.)

Question: Can you guess why my

students decided to refer to this result

as the “Opera House” theorem?

Actually more is true in this famous

theorem:

... and common measure of the

peripheral angles is half the measure of

the arc: 1

2 x y= .

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The measure of an arc is simply the

“amount of turning” it represents. As an

actual angle it can be found as the angle

between the two radii of the circle that

reach the endpoints of the arc.

Now the challenge: How might we prove

that the measure x is half of the

measure y in the picture below?

It seems compelling to draw in a third

radius. This creates for us isosceles

triangles, whose congruent base angles

I’ve labeled a and b .

We see x a b= + .

Can we get a formula for y in terms of

a and b ? You bet!

Look at the three angles around the

center point of the circle. These sum to

a full 360 o . One of these angles is y ,

another is 180 2a− , and the third is

180 2b− . We thus have:

180 2 180 2 360y a b+ − + − = .

This gives 2 2y a b= + and, indeed,

x a b= + is half of this.

Exercise: The picture we drew was too

nice. Show that 1

2 x y= in this lopsided

picture too!

The Opera House theorem has some

lovely consequences:

Thales’ Theorem: The angle subtended

from a diameter of a circle is a right

angle.

Question: Thales’ (ca. 624 – ca. 546

BCE), the “father of geometry,” did not

use the Opera House theorem to

establish this result. What is a much

simpler away to prove Thales’ theorem?

(Hint: Draw in one radius.)

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Cyclic Quadrilaterals: Opposite angles

of a quadrilateral inscribed in a circle

sum to half a turn.

Question: According to the Opera

House theorem, why is 2 2x y+ one full

turn?

Comment: Since the four angles in any

quadrilateral sum to 360 o , if one pair of

opposite angles are supplementary,

then so too is the remaining pair.

THE OPERA HOUSE THEOREM

CONVERSE?

If you tried the second paper-pushing

experiment it seems that the tip of the

paper again traces the arc of a circle.

Now we know from the Opera House

theorem that points on the arc of a

circle that passes through the two nails

subtend the same angle to those two

nails.

We are now pursuing the converse:

If a curve has the property that points

on it subtend the same angle from two

fixed points, must that curve be part of a

circle?

Here’s how to prove this is so.

STEP 1: Any three non-collinear points

are con-cyclic.

First note that any point on the

perpendicular bisector of the line

segment connecting two points A and

B is equidistant from A and B . This is

a consequence of the Pythagorean

theorem.

2 2

a x h b= + =

Given three non-collinear points, the

point where two perpendicular

bisectors intersect is equidistant from

all three points.

This intersection point is thus the center

of a circle that passes through all three

of the given points.

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STEP 2: Place the paper up between the

two nails one time and draw the circle

that passes through the tip of the paper

and the two nails.

In this picture, x is half of y .

STEP 3: We need to show that if we

reinsert the paper up between the two

nails a second time, its tip is sure to land

on the same circle.

Let’s suppose it doesn’t and see what

goes wrong. There are two cases to

consider.

Suppose the tip of the paper lands

outside the circle.

If we draw the dashed line shown to

highlight angle m , we see, by the Opera

House theorem that 1

2 m y= , just like

x . This is suspicious!

Draw the chord connecting the two

nails. We now have two triangles with

the following angle configurations.

We see:

180b c x+ + = o

180a b c x+ + + = o

giving 180 180a+ = o o

. Oops!

This shows that it can’t be the case that

the tip of the paper lands outside the

circle.

Exercise: Show that a contradiction also

arises if we assume the tip of the paper

lands at a position inside the circle.

We can only conclude that whenever

we reinsert the paper, its tip is sure to

land on the same circle. The curve

traced by that tip is that circle!

COOL TIP: Suppose you need to find the

exact radius of a flower pot. Lay a piece

of paper across it as shown. You have

now marked off an exact diameter!

Question: How can you use paper, a

marker, and string to find the exact

center of the pot?

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By the way … We are all set to prove

the converse of the Cyclic Quadrilateral

theorem too:

Theorem: Suppose a quadrilateral has

opposite angles that are supplementary.

Then that quadrilateral is cyclic (that is,

all four vertices of that quadrilateral sit

on a common circle.)

[Draw the circle that passes through

three vertices of the quadrilateral. Why

must the fourth vertex sit on that circle

as well?]

We’ll make good use of this final result

in this month’s COOL MATH ESSAY.

© 2014 James Tanton

[email protected]