Difference between revisions of "Radical axis"

Revision as of 21:08, 5 June 2014

Introduction

The theory of radical axis is a priceless geometric tool that can solve formidable geometric problems fairly readily. Problems involving it can be found on many major math olympiad competitions, including the prestigious USAMO. Therefore, any aspiring math olympian should peruse this material carefully, as it may contain the keys to one's future success.

Definitions

The power of point $A$ with respect to circle $\omega$ (with radius $r$ and center $O$ , which shall thereafter be dubbed $pow(P, \omega)$ , is defined to equal $OP^2 - r^2$ .

Note that the power of a point is negative if the point is inside the circle.

The radical axis of two circles $\omega_1, \omega_2$ is defined as the locus of the points $P$ such that the power of $P$ with respect to $\omega_1$ and $\omega_2$ are equal. In other words, if $O_i, r_i$ are the center and radius of $\omega_i$ , then a point $P$ is on the radical axis if and only if $PO_1^2 - r_1^2 = PO_2^2 - r_2^2$

Results

Theorem 1: (Power of a Point) If a line drawn through point P intersects circle $\omega$ at points A and B, then $|pow(P, \omega)| = PA * PB$ . Theorem 2: (Radical Axis Theorem)

a. The radical axis is a line perpendicular to the line connecting the circles' centers (line $l$ ).

b. If the two circles intersect at two common points, their radical axis is the line through these two points.

c. If they intersect at one point, their radical axis is the common internal tangent.

d. If the circles do not intersect, their radical axis is the perpendicular to $l$ through point A, the unique point on $l$ such that $pow(A, \omega_1) = pow(A, \omega_2)$ .

Theorem 3: (Radical Axis Concurrence Theorem) The three pairwise radical axes of three circles concur at a point, called the radical center.

Proofs

Theorem 1 is trivial Power of a Point, and thus is left to the reader as an exercise. (Hint: Draw a line through P and the center.)

Theorem 2 shall be proved here. Assume the circles are $\omega_1$ and $\omega_2$ with centers $O_1$ and $O_2$ and radii $r_1$ and $r_2$ , respectively. (It may be a good idea for you to draw some circles here.)

First, we tackle part (b). Suppose the circles intersect at points $X$ and $Y$ and point P lies on $XY$ . Then by Theorem 1 the powers of P with respect to both circles are equal to $PX * PY$ , and hence by transitive $pow(P, \omega_1) = pow(P, \omega_2)$ . Thus, if point P lies on $XY$ , then the powers of P with respect to both circles are equal.

Now, we prove the inverse of the statement just proved; because the inverse is equivalent to the converse, the if and only if would then be proven. Suppose that P does not lie on $XY$ . In particular, line $PY$ does not intersect X. Then $PY$ intersects circles $\omega_1$ and $\omega_2$ a second time at distinct points $M$ and $N$ , respectively. (If $PY$ is tangent to $\omega_1$ , for example, we adopt the convention that $P = M$ ; similar conventions hold for $\omega_2$ . Power of a Point still holds in this case. Also, notice that $M$ and $N$ cannot both equal $P$ , as $PY$ cannot be tangent to both circles.) Because $PM$ is not equal to $PN$ , $PY \cdot PM$ does not equal $PY \cdot PN$ , and thus by Theorem 1 $pow(P, \omega_1)$ is not congruent to $pow(P, \omega_2)$ , as desired. This completes part (b).

For the remaining parts, we employ a lemma:

Lemma 1: Let $P$ be a point in the plane, and let $P'$ be the foot of the perpendicular from $P$ to $O_1O_2$ . Then $pow(P, \omega_1) - pow(P, \omega_2) = pow(P', \omega_1) - pow(P', \omega_2)$ .

The proof of the lemma is an easy application of the Pythagorean Theorem and will again be left to the reader as an exercise.

Lemma 2: There is an unique point P on line $O_1O_2$ such that $pow(P, O_1) = pow(P, O_2)$ .

Proof: First show that P lies between $O_1$ and $O_2$ via proof by contradiction, by using a bit of inequality theory and the fact that $O_1O_2 > r_1 + r_2$ . Then, use the fact that $O_1P + PO_2 = O_1O_2$ (a constant) to prove the lemma.

Lemma 1 shows that every point on the plane can be equivalently mapped to a line on $O_1O_2$ . Lemma 2 shows that only one point in this mapping satisfies the given condition. Combining these two lemmas shows that the radical axis is a line perpendicular to $l$ , completing part (a).

Parts (c) and (d) will be left to the reader as an exercise. (Also, try proving part (b) solely using the lemmas.)

Now, try to prove Theorem 3 on your own! (Hint: Let P be the intersection of two of the radical axises.)

Exercises

If you haven't already done so, prove the theorems and lemmas outlined in the proofs section. If you have done so, prove them again. (Repetition is the best way to master a new concept.)

Note: No solutions will be provided to the following problems. If you are stuck, ask on the forum.

Problem 1. Two circles P and Q intersect at X and Y. Point A is located on $XY$ such that AP = 10 and AQ = 15. If the radius of Q is 7, find the radius of P.

Problem 2. Solve 2009 USAMO Problem 1. If you already know how to solve it, write up a real proof on a piece of paper. And do it twice.

Problem 3. Two circles P and Q with radii 1 and 2, respectively, intersect at X and Y. Circle P is to the left of circle Q. Prove that point A is to the left of $XY$ if and only if $AQ^2 - AP^2 > 3$ .

Problem 4. Solve 2012 USAJMO Problem 1.

@@ Line 41: / Line 41: @@
 ''Lemma 2:'' There is an unique point P on line <math>O_1O_2</math> such that <math>pow(P, O_1) = pow(P, O_2)</math>.
-Proof: First show that P lies between <math>O_1</math> and <math>O_2</math>, by using a bit of inequality theory and the fact that <math>O_1O_2 > r_1 + r_2</math>. Then, use the fact that <math>O_1P + PO_2 = O_1O_2</math> (a constant) to prove the lemma.
+Proof: First show that P lies between <math>O_1</math> and <math>O_2</math> via proof by contradiction, by using a bit of inequality theory and the fact that <math>O_1O_2 > r_1 + r_2</math>. Then, use the fact that <math>O_1P + PO_2 = O_1O_2</math> (a constant) to prove the lemma.
 Lemma 1 shows that every point on the plane can be equivalently mapped to a line on <math>O_1O_2</math>. Lemma 2 shows that only one point in this mapping satisfies the given condition. Combining these two lemmas shows that the radical axis is a line perpendicular to <math>l</math>, completing part (a).

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