Mathematical Problem Solving Power of a Point A rather simple definition of the power of a point with respect to a circle is: Let C be a circle of radius r. The power of a point P with respect to C is given by d2 − r2, where d is the distance of P to the center of the circle. Just to fix ideas, for example, if C is the circle of radius r centered at the origin, and P has coordinates (x, y), then the power of P with respect to this circle is x2 + y2 − r2. If P is a point, C a circle, I’ll write Π(P, C) to denote the power of P with respect to C.

Obviously, Π(P, C) > 0 if and only if P is outside the circle, < 0 if and only if P is inside the circle, and 0 if and only if P is on the circle.

The significance of this concept is due to the following result. Theorem 1 Let P,X,X0 be collinear points, let C be a circle. If X,X0 lie on C, and either X 6= X0, or if X = X0 6= P and the line containing X and P is tangent to C, then 0 Π(P, C)= PX · PX , where PX,PX0 are “directed lengths;” to be specific: PX · PX0 < 0 if P is strictly between X,X0, > 0 if either X is strictly between P and X0 or X0 is strictly between P and X, 0 if P coincides with X or X0. To clarify a bit more this “directed lengths” thing, P divides the line containing P,X,X0 into two parts or sides. Call one part the positive part the other side the negative part. If Z is a point on the line, define PZ to be the length of the segment from P to Z if Z is on the positive side, to be minus that length if Z is on the negative side. Which side is chosen to be positive or negative is wholly arbitrary. Proof. Before beginning the actual proof, we need to get some trivialities out of the way. The case in which X = P or X0 = P is trivial. This case is equivalent to PX · PX0 = 0. Obviously, if one of X or X0 is P , then X,X0 are on a circle C if and only if P is on the circle; that is, if and only if ¶(P, C) = 0. From now on, we may assume that X 6= P and that X0 6= P . We will assume first that X 6= X0. If P is not between X and X0, then X,X0 are on the same ray emanating from P ; we can 0 then set up a system of cartesian coordinates with P at the origin and write X = (r1 cos θ, r1 sin θ), X = (r2 cos θ, r2 sin θ), 0 where r1, r2 > 0; r1 6= r2, θ ∈ [0, 2π). The fact that both X,X have the same second coordinate θ is because they are collinear with the origin P . By taking the direction along the ray of angle θ with the positive x-axis as the positive direction 0 0 along that ray, we’ll have PX = r1, PX = r2. On the other hand if P is between X and X we can also set up cartesian 0 coordinates with P at the origin and write X = (r1 cos θ, r1 sin θ), X = (r2 cos θ, r2 sin θ) for some th ∈ [0, 2π), but this time 0 one of r1, r2 is positive, the other one negative; say r1 > 0, r2 < 0. This time PX · PX = r1r2 < 0. Suppose that C is the circle of center (a,b), radius R. That X is on C implies

2 2 2 (r1 cos θ − a) + (r1 sin θ − b) = R , and this expands to (using cos2 θ + sin2 θ = 1):

2 2 2 2 r1 − 2(a cos θ + b sin θ)r1 + (a + b − R )=0. Similarly, X0 on C implies 2 2 2 2 r2 − 2(a cos θ + b sin θ)r2 + (a + b − R )=0.

This implies that r1, r2 are roots of the quadratic equation r2 − 2(a cos θ + b sin θ)r + (a2 + b2 − R2) = 0;

2 2 2 since they are distinct, their product r1r2 must equal the constant term of this equation; i.e., r1r2 = a + b − R . But 0 2 2 2 2 2 r1 = PX. r2 = PX and a + b − R = Π(P, C); the latter because a + b is the square of the distance of the center of C to the origin, which is P . This proves: If X 6= X0 on C, then PX · PX0 = Π(P, C). Assume now that the line from P to X is tangent to C at X. If O denotes the center of the circle, then the triangle OPX is a right triangle of hypotenuse P O, thus P O2 = PX2 + XO2. Now P O is the distance from P to the center of the circle, XO is the radius of the circle, thus P O2 − XO2 = Π(P, C) and we proved that Π(P, C)= PX2in this case.

There is a converse of sorts to this theorem. 0 0 0 0 0 0 Theorem 2 Let P,X,X ,Y,Y be points in the plane, assume X 6= X , Y 6= Y , P,X,X lie on a line `1 and that P,Y,Y 0 lie on a line `2 6= `1. (Another way of stating this assumption is to say that the line determined by X,X intersects the line determined by Y, Y 0 at P .) Then there exists a circle C containing X,X0,Y,Y 0 if and only if PX · PX0 = P Y · P Y 0. In this case, we also have Π(P, C)= PX · PX0 = P Y · P Y 0.

Proof. If C is a circle such that X,X0,Y,Y 0 lie on C, then by Theorem 1 PX · PX0 = Π(P, C)= P Y · P Y 0. For the converse assume that PX · PX0 = P Y · P Y 0. 0 0 0 0 ∗ Conversely, assume that PX · PX = P Y · P Y . Since X,X , Y are not collinear (X,X are on `1 and Y is not) , there is a circle through X,X0, Y . The picture now looks somewhat like the picture below. I am assuming for the picture, but not for the proof, that P lies outside of the circle.

0 X b

X b

Y’ b b P b Y

A line can intersect a circle zero, one or two times. The line containing P, Y already intersects the circle once, at Y . If the is the only intersection then the line is tangent to the circle; set Z = Y . Otherwise, let Z be the other intersection of the line with the circle.

0 X b

X b

Y’ b b b Z P b Y

By Theorem 1, the (direct) power of a point theorem, PX · PX0 = P Y · PZ. But we also have PX · PX0 = P Y · P Y 0, hence P Y · P Y 0 = P Y · PZ. Since P Y 6= 0, this implies P Y 0 = PZ. The last equality is only possible if Y 0 = Z, given that P,Y,Y 0,Z are all on the same line and PY,PY 0,PZ are really coordinates on that line with P at 0. That means that Y 0 is on the circle determined by X,X0, Y .

1 Ceva’s Theorem

Theorem 3 (Ceva) Let ABC be a triangle. Let D,E,F be points on the sides BC,CA, and AB, respectively. The segments (cevians) joining A to D, B to E and C to F are concurrent if and only if the following relation holds:

AF BD CE · · =1. FB DC EA We will prove this theorem using some basic properties of complex numbers. But, first here is a picture of the situation.

∗There is a pesky trivial possibility; Y could be P . Then PY · PY 0 = 0 and PX · PX0 = PY · PY 0 implies PX · PX0 = 0 so one of X,X0 must be P , say X = P . So we don’t have four points X,X0,Y,Y 0 but only three, X,Y,Y 0 and since there always is a circle through any three non-collinear points, there is nothing further to prove.

2 A

E b b F

C b B D

Here is the picture with the cevians drawn in.

A

E b b F

C b B D

These cevians are obviously not concurrent.

We will interpret the vertices A, B, C as complex numbers, placing C at the origin, so C = 0, B on the positive real axis (so B is a real number), and A wherever it has to go. We may assume that A is in the upper half plane, though not necessarily in the first quadrant. We need to know some simple facts about the geometry of complex numbers. 1. Let α, β ∈ C. The line segment from α to β consists of all complex numbers of the from (1 − s)α + sβ, where s is a real number, 0 ≤ s ≤ 1. For example, the line segment from the origin to the point 1 + i consists of all numbers of the form (1 − s)0 + s(1 + i) = s + si, where 0 ≤ s ≤ 1. As another example, suppose that α, β are real.It is then aneasy exercise to see that a complex number z is of the form (1 − s)α + sβ if and only if z is real and α ≤ z ≤ β. For s = 0 one gets α, when s = 1, one gets β, and when s =1/2 one gets (α + β)/2, the midpoint of the segment. Notice: (1 − s)α + sβ = α + s(β − α).

3 2. Let us begin writing complex numbers using Latin capitals, to prepare ourselves for the proof to come. Suppose A, B are complex numbers and C is on the line segment joining A to B. If C = (1 − s)A + sB, then

AC s = . CB 1 − s (By AC we understand the length of the segment from A to C; etc.) In fact, we have:

AC = |C − A| = |(1 − s)A + sB − A| = |A − sA + sB − A| = |s(B − A)| = s|B − A|, CB = |B − C| = |B − (1 − s)A − sB| = |B − A + sA − sB| = |(1 − s)(B − A)| = (1 − s)|B − A|,

and the result follows. Notice that |sα| = s|α|, |(1 − s)α| = (1 − s)|α| for a complex number α because s, 1 − s ≥ 0. 3. If A, B, C are the vertices of a triangle so that A, B, C are not collinear, then A − C,B − C are linearly independent over the reals. That means that if s,t,u,v are real numbers such that

s(A − C)+ t(B − C)= u(A − C)+ v(B − C),

then s = u and t = v. We are ready for the proof. At least, I am ready. Are you? Proof. As mentioned above, we set up coordinates so that C = 0 and B is a positive real number. As vectors, A = A − C and B = B − C are linearly independent. Since D is on the segment from 0 to B, we can write D = sB, where s is a real number, 0

A

E b b F Q b

C b B D

4 The point Q is on the line segment from A to D, thus there is a real number v, 0

Q = (1 − v)A + vD = (1 − v)A + vsB.

The point Q is also on the line segment from B to E, thus there is a real number w, 0

Q = (1 − w)B + wE = (1 − w)B + wtA.

Because A, B are linearly independent, we must have 1 − v = tw and 1 − w = sv. We can solve here for v, w to get 1 − t 1 − s v = , w = . 1 − st 1 − st Getting finally to the real proof, let us assume first that Q is also on the line segment from C =0 to F = (1 − u)A + uB. Then there is a real number x such that 0

A

E b b F b 0 Q F b

C b B D

By the direct part of the proof, we must have

AF 0 BD CE · · =1. F 0B DC EA

5 On the other hand, we are assuming AF BD CE · · =1, FB DC EA thus AF AF 0 = . FB F 0B With F, F 0 points on the line segment from A to B, this equality of ratios is only possible if F = F 0; that is, the line segment from A to F is concurrent at Q with the other two cevians. Do you see the similarity with the proof of the converse of the Power of a Point Theorem (Theorem 2)?

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