MATH 5200/7200 Exam 3 Solutions

1. (a) What is the relation between the sine of an angle in a triangle, the length of the opposite side, and the circumradius? (Do not prove this
relation.)

This is a version of the law of sines:

a / (sin A) = 2R

For the triangle ABC, sin A stands for (the absolute value of) the sine of angle CAB, a is the length of BC, and R is the radius of the circumcircle:

Comment: You can check this formula by applying it to a right triangle.

(b) State Ptolemy's theorem. (Do not prove it.)

Ptolemy's theorem: For a cyclic quadrilateral (a quadrilateral which can be inscribed in a circle) the sum of the products of the lengths of the
opposite sides equals the product of the lengths of the diagonals.

Comment: It's important that you multiply opposite sides, not adjacent sides.

(c) Use parts (a) and (b) and the following diagram to derive the formula for the sine of the sum of two acute angles x and y.

This quadrilateral has supplementary opposite angles (the two right angles), so it is a cyclic quadrilateral. Ptolemy's theorem says that ac + bd = mn.

By hypothesis m = 1. By the definition of sine, sin(x) = sin(CAB) = d/m = d, and sin(y) = sin(DAC) = a/m = a. By the definition of cosine, cos(x) =
cos(CAB) = c/m = c, and cos(y) = cos(DAC) = b/m = b.

We can apply part (a) to angle DAB of the triangle ABD. The angle measure a(DAB) = x + y. by the angle addition axiom. The circumcircle of
ABD is also the circumcircle of the right triangle ABC, so AC is a diameter of this circle. Hence the circumradius of ABD is m/2 = 1/2. Thus part
(a) implies that sin(x+y) = n.

Therefore Ptolemy's theorem translates to

sin(y) cos(x) + cos(y) sin (x) = 1 sin(x+y),

sin(x+y) = sin(x) cos(y) + cos(x) sin(y).

Comment: To apply Ptolemy's theorem you must first show that the quadrilateral ABCD is cyclic.

2. Use the law of sines to prove the following theorem from exam 2:

The bisector of an angle of a triangle divides the opposite side into segments that are proportional to the adjacent sides. In other words, in the
diagram below if AD is the bisector of angle A, then BD/DC = BA/AC. Hint: Apply the law of sines to the triangles ABD and ACD.

Proof:

Let a be the angle measure of angle BAD. Since AD bisects angle BAC, a is also the angle measure of angle DAC. Let x be the angle measure of
angle ADB, and let y be the angle measure of angle CDA.

By the law of sines applied to the triangle ABD, BD/sin a = BA/sin x.

By the law of sines applied to the triangle ACD, DC/sin a = AC/sin y.

Therefore BD/DC = (BA/AC)(sin y/sin x).

Since x and y are supplementary, sin x = sin y. Therefore BD/DC = BA/AC, QED.

Comment: The law of sines can not be applied to angles from two different triangles! For example it is not true that AD/sinB = AD/sinC, since ABD and ACD are not the same triangle.

3. Definition: Two circles are tangent externally at a point if the circles pass through the point, have the same tangent line at the point, and lie on
opposite sides of this common tangent line.

Suppose that three circles X, Y, and Z with centers P, Q, and R are tangent externally at points A, B, and C, and the chords AB of X and and AC of
Y are extended to intersect Z at points D and E. Prove that D, R, and E are collinear.

Hint: Show angles PAB and RDB are equal and angles CAQ and CER are equal.

Proof:

Let T be the common tangent line of circles X and Y at the point A. We know that the radius PA of X is perpendicular to T at A, and the radius QA
of Y is perpendicular to T at A. By uniqueness of perpendiculars, these radii lie on the same line, i.e. the points P, A, and Q are collinear. The same
argument shows that the points P, B, and R are collinear, and the points Q, C, and R are collinear.

Since PA and PB are radii of the circle X, PA = PB, so the triangle PAB is isosceles. By the isosceles triangle theorem the angles PAB and ABP
are equal. The same argument shows that triangle QAC is isosceles, so angles CAQ and QCA are equal. Similarly, triangle RDB is isosceles, so
angles RDB and DBR are equal, and triangle RCE is isosceles, so angles CER and RCE are equal.

Now angles ABP and DBR are equal, by the vertical angle theorem. (Here we use that P, B, and R are collinear.) Similarly, angles QCA and RCE
are equal. (Here we use that Q, C, and R are collinear.)

Since a(PAB) = a(ABP), a(ABP) = a(DBR), and a(DBR) = a(RDB), we have

a(PAB) = a(RDB).

Since a(CAQ) = a(QCA), a(QCA) = a(RCE), and a(RCE) = a(CER), we have

a(CAQ) = a(CER).

Now angles PAB and RDB are alternate interior angles for the transversal AD of the lines PA and DR, so PA and DR are parallel. Also, angles
CAQ and CER are alternate interior angles for the transversal AE of the lines QA and ER, so QA and ER are parallel. Now P, A, Q are collinear,
so the lines DR and ER are parallel to the line PQ, and both pass through the point R. By the uniqueness of parallels, the lines DR and ER are
equal, so the points D, R, and E are collinear, QED.

[Here is an alternate last paragraph for the proof:

Since P, A, Q are collinear, angle PAQ is a straight angle, so a(PAB) + a(BAC) + a(CAQ) = 180. Therefore, by substitution a(RDB) + a(BAC) +
a(CER) = 180, or a(RDA) + a(DAE) + a(AER) = 180. Since the sum of the angles of the quadrilateral ADRE is 360, it follows that a(ERD) = 180,
so D, R, E are collinear by the straight angle theorem, QED.]

Comment: The key to this proof is to use the hypothesis that the three circles are tangent. In the above proof this hypothesis is use to prove that P, A,
Q are collinear, P, B, R are collinear, and Q, C, R are collinear.