MATH 5200/7200 Homework 10 solutions
1. Suppose that two lines through a point P
are tangent to a circle at points A and B. If the circle has center
O and radius R, and Q is the intersection
point of the lines OP and AB, then (OP)(OQ) = R2.
The triangles OAP and OBP have the side OP
in common. The sides OA and OB are radii of the circle, so they
each have length R. Since AP is
tangent to the circle at point A, angle OAP is a right angle.
Similarly, angle OBP is a right angle. By the Pythagorean theorem,
AP is the square root
of (OP)2 - (OA)2, and BP is the square root of (OP)2 - (OB)2;
thus AP = BP. Therefore triangles OAP and OBP are congruent by
the side-side-side
congruence theorem. Thus the corresponding angles AOP and BOP
are congruent. (Since A and B lie on opposite sides of the line
OP, a(AOP) =
-a(BOP), and so the triangles OAP and OBP are negatively congruent.)
Therefore a(AOQ) = -a(BOQ), and so the triangles
OAQ and OBQ are negatively congruent by the side-angle-side congruence
theorem. Thus
a(AQO) = -a(BQO), so a(AQO) = a(OQB). Since a(AQO) + a(OQB) =
180, the angles AQO and BQO are right angles.
Since a(POA) = a(QOA), the right triangles
PAO and AQO are negatively congruent, by the angle-angle congruence
theorem. Therefore OP/OA =
OA/OQ, so (OP)(OQ) = (OA)2 = R2.
2. If two lines through a point P meet a circle
at points A, A' and B, B', respectively, then (PA)(PA') = (PB)(PB').
(Explanation: One line meets the
circle at two points A and A', and the other line meets the circle
at two points B and B'.)
There are two cases, when P is outside the circle and when P is inside the circle. The same proof works in both cases!
First note that a(B'PA) = a(BPA) = -a(APB)
= -a(A'PB). Now a(AB'B) = a(AA'B) by the inscribed angle theorem,
so a(AB'P) = a(AB'B) =
a(AA'B) = a(PA'B) = -a(BA'P)
Therefore the triangles PAB' and PBA' are negatively similar by the angle-angle similarity theorem.
Thus PA/PB = PB'/PA', so (PA)(PA') = (PB)(PB'), QED.
Note: A version of this theorem is also true
if the points A, A', B, B' are not distinct. The most interesting
of these cases is when A = A', and PA is
tangent to the circle, but B is not equal to B':
Then triangles PAB' and PBA are negatively similar, so PA/PB = PB'/PA, so (PA)2 = (PB)(PB').
3. The feet of the perpendiculars from a point
to the sides of a triangle are collinear if and only if the point
lies on the circumcircle. (Explanation:
The foot of the perpendicular from a point to a line is the intersection
of the line with the perpendicular line through the point.)
There are several slightly different cases to consider, but for simplicity we'll consider only the case illustrated by the following sketch:
Let ABC be a triangle, and let P be a point
Let X be the foot of the perpendicular from P to BC, Y the foot
of the perpendicular from P to AC, and Z
the foot of the perpendicular from P to AB. We must prove that
P lies on the circum circle of ABC if and only if X, Y, Z are
collinear.
First we assume that P lies on the circumcircle
of ABC, and we prove that XYZ are collinear. Consider the angles
labelled in the figure: 1 = CYX, 2
= CPX, 3 = XCP, 4 = PCB, 1' = AYZ, 2' = APZ, 3' = ZAP. We will
prove that angle 1 equals angle 1', which gives that X, Y, Z are
collinear.
Since angles CYP and CXP are right angles,
the quadrilateral YCXP is cyclic, so angle 1 equals angle 2, since
they both subtend the chord CX of
the circumcircle of YCXP. Similarly, the quadrilateral AZYP is
cyclic, so angle 1' equals angle 2'. Now angles 2 and 3 are complementary,
since
CXP is a right angle. Similarly angles 2' and 3' are complementary.
Also angles 3 and 4 are supplementary. But since ABCP is a cyclic
quadrilateral by hypothesis, angles 3' and 4 are supplementary.
Therefore angle 3 equals angle 3', and it follows that angle 1
= angle 1'.
To prove the converse, assume X, Y, Z are collinear.
We'll prove that P lies on the circumcircle of ABC, i.e. that
the quadrilateral ABCP is cyclic.
Since X, Y, Z are collinear, angles 1 and 1' are equal. We can
still use the facts that YCXP and AZYP are cyclic to get that
1 = 2 and 1' = 2'. We also
have that 2 and 3 are complementary, that 2' and 3' are complementary,
and that 3 and 4 are supplementary. Putting this all together,
we conclude
that angles 3 and 3' are supplementary. It follows that ABCP is
cyclic, QED.