Solutions to the Midterm Exam
Math 5200/7200
March 24, 2008
1. Recall that two circles c1and c2are mutually tangent at Tif there is a line lthrough Tsuch that lmeets both
c1and c2tangentially at T.
(a) (5 points) Let c1with center O1and c2with center O2be mutally tangent at T. Show that the line
segment O1O2is perpendicular to the common tangent line l.
Pick P∈lwith P6=T. Then ∠PTO1=π/2 and ∠PTO2=π/2, which implies O1, T, O2are collinear.
This means lmeets O1O2perpendicularly.
(b) (5 points) Show that mutually tangent circles c1and c2can only meet at one point. (Hint: Use part (a)
to derive a contradiction if they meet at second point.)
Suppose c1and c2meet at some point P6=T. By the definition of tangent line (that lcan only meet c1at
one point), P6∈ l. Because P6∈ l,∠PTO16=π/2, so either ∠PTO1< π/2 or ∠PTO1> π/2. Notice that
|T O1|=|P O1|, so ∠PTO1=∠TPO1. Thus, if ∠PTO1> π/2, then ∆PTO1has an angle sum greater
than π, which is impossible. Next observe that ∠PTO1and ∠PTO2are supplementary, so ∠PTO1< π/2
implies ∠PTO2> π /2, so the isosceles triangle ∆PTO2has angle sum greater than π. This contradiction
completes the proof that there can only be one intersection point between c1and c2.
2. We showed in one of the homework assignments that, given a line l, there are constant A, B, C (with one of
A, B nonzero) such that (x, y)∈lif and only if Ax +By =C.
(a) (5 points) What is the relationship between the vector (A, B) and l? (Hint: pick a basepoint and use the
dot product.)
Let (x, y)∈l, then (A, B)·(x, y ) = Ax +By =C. Thus, choosing a base point (x0, y0), we see
0 = C−C= (A, B)·(x, y)−(A, B )·(x0, y0)=(A,B )·(x−x0, y −y0).
In other words (A, B) is perpendicular to any displacement vector (x−x0, y −y0) on the line l. We
conclude (A, B) is perpendicular to the line.
(b) (5 points) What is the condition on the coefficients A, B, C such that lpasses through the origin (0,0)?
Suppose lpasses through (0,0). Then C=A·0 + B·0 = 0. Conversely, suppose C6= 0. Then
A·0 + B·0=06=C, so (0,0) cannot lie on l. We conclude (0,0) ∈lif and only if C= 0.
3. Let ABCD be a square.
(a) (6 points)If lis the perpendicular bisector of AB, show that the reflectionrlmaps the square to itself.
By the definition of reflection, rl(A) = B. Let lpass through AB at X, and let lpass through DC
at Y. Then |AX|=|BX|,|XY |=|X Y |, and ∠AXY =∠BX Y , so ∆AXY = ∆BX Y by SAS. Then
|AD|=|BC |,|AY |=|BY |and ∠ADY =∠BCY =π/2, so ∆ADY = ∆B CY by hypotenuse-leg for
right triangles. Thus |DY |=|C Y |and ∠DY X =∠CY X =π/2, and so lis the perpendicular bisector of
DC and rl(D) = C. We’ve shown rlmape vertices of ABCD to other vertices of ABCD, so rlmaps
the square to itself.
(b) (6 points) If mis the angle bisector of ∠DAB, show that the reflection rmmaps the square to itself.
First note ∆ADC = ∆ABC by SSS, so ∠DAC =∠B AC and ∠DCA =∠BCA, which means the diagonal
AC is the angle bisector mof both ∠DAB and ∠DC B. Because A, C are both on m, we have rm(A) = A
and rm(C) = C. Now let DB cross mat O. Then |DA|=|BA|,|AO|=|AO|and ∠DAO =∠BAO, so
∆DOA = ∆BOA. Then |DO|=|BO|and ∠DOA =∠BOA =π /2, do mis the perpendicular bisector
of DB. Thus rm(B) = D.
4. (6 points) Recall that a glide reflection Gl,AB =TAB ◦rlis the composition of a relfection rland a translation
TAB. Show the composition of Gl,AB with itself, that is G2
l,AB, is the translation T2AB . (Hint: rewrite one of
the copies of Gl,AB.)
Recall that we can write a glide reflection as
Gl,AB =TAB ◦rl=rl◦TAB,
so
G2
l,AB =TAB ◦rl◦rl◦TAB =TAB ◦TAB =T2AB.
