VMHS Math Circle
XII. The Grand Geometry Review (Part II: Synthetic Geometry-Circles)
Circles are intriguing. To me, it’s because they’re associated with pi (π), a number that seems so
dubious. π2/6 is the sum of the infinite series (1/1) + (1/22) + (1/32) + … + (1/n2) + … Pi is just so
interesting. However, what this document will cover are certain formulas and tricks you should know in
order to deal with AMC problems dealing with circles. This won’t be as extensive as our lines and
triangles review, but make sure you know what’s going to come.
Formulas
Basic Circumference/Area:
Circumference = 2πr, with “r” being the radius
Area = πr2, with “r” being the radius
Arcs and Sectors:
Arc Length = 2πr(n˚/360˚), with “r” being the radius and “n” being the degree measure of the central
angle subtending this arc (think of subtending as cutting into an arc).
Sector Area = πr2(n˚/360˚), with “r” being the radius and “n” being the degree measure of the central
angle subtending this arc.
n˚
Two-Tangent Theorem:
Consider a point P outside a circle C. If PA and PB are segments both tangent to the C, they have the
same length. If you don’t know what a tangent line is, it’s a line that intersects a curve/shape/circle at only
one point.
A
P
C
B
Secants, Tangents, Other Lines, and Angles:
If a circle’s radius and a tangent line intersect, the angle formed by the intersection is 90˚.
If you have two secants (lines intersecting a shape/curve/circle at two points), and both secants meet at a
point P lying on the circle, then the central angle is twice the measure of the angle determined by those
secants.
2n˚
n˚
P
There are a lot of other cases. You don’t need to worry about this for the AMC, but it might help. Refer to
the diagrams below for formulae. Note that “x” denotes the measure of the larger angles/arcs while “y”
denotes the smaller.
((x + y)/2)˚
y˚
((x – y)/2)˚
((x – y)/2)˚
x˚
y˚
x˚
x˚
y˚
((x – y)/2)˚
y˚
x˚
Power of a Point:
If you have two chords AB and CD in a circle intersecting at a point R, (AR)(BR) = (CR)(DR).
If you have a tangent AR and a secant RC (passing through point B on the circle) meeting at point R
outside the circle, AR2 = (RB)(RC).
If you have two secants RB and RY (passing through A and X on the circle, respectively) meeting at point
R outside the circle, (RA)(RB) = (RX)(RY).
A
D
A
B
R
A
R
C
B
B
R
X
C
Y
Ptolemy’s Theorem:
If you have a cyclic quadrilateral (a quadrilateral that can be inscribed in a circle) ABCD, with sides AB,
BC, CD, and DA, along with diagonals AC and BD, then (AB)(CD) + (BC)(DA) = (AC)(BD).
A
B
D
C
Know that the opposite angles of a cyclic quadrilateral add up to 180 degrees!
To be honest, this is basically all you need to know about circles. Ptolemy’s theorem will likely
be featured in the final five problems. Let’s get to some practice. Problems here will be presented in
increasing difficulty. Note that quite a few of these will involve overlap with what we learned about lines
and triangles.
The strip of a 2-d conveyor belt is held by four congruent circles, each with radius 4. Assuming that it
fully wraps around all four circles, what is the length of the strip of this conveyor belt?
Okay, let’s do some cutting.
From the diagram, we basically need to count the circumference of two semicircles, plus six diameters
(not three because we need to consider both the top and bottom part of the blue section). Thus, we get
2π(4) + 6(8) = 8π + 48.
The logo for Mitsubishi Motors is a red equilateral triangle but with equilateral triangles cut off from the
middle portion of each of its sides when those sides are trisected. If the sides of the original equilateral
triangle have length 1, what is the area swept out by a 60˚ rotation clockwise of the logo?
Now, remember that finding the area that is “swept out” means that if we pick up the Mitsubishi logo,
paint one side of it entirely, put it down with the painted side on the ground, spin it 60˚, and pick it back
up again, we want to find the area of the painted region. Let’s do that accordingly.
It is a bit tough to visualize how to find the area of this shape. However, let’s draw some lines and
enclose this figure around a full circle.
Now we know what we’re dealing with. We know how to find the area of an equilateral triangle, as well
as the areas of those mini-ones. However, finding the circle-sector-esque regions seems a bit tougher.
What we can do is take the circle’s area, subtract it by the areas of the four equilateral triangles seen
here, and divide it by 2 (just visualize this and you’ll get it… I hope). Doing so, we have to add this to the
areas of our four triangles. Note that the radius to find the area of the circle is (1/2)/√(3) * 2 = √(3)/3.
Hence, we get:
[π [√(3)/3]2 – (1 * √(3)/4 + √(3)/6 * 1/2 * 1/3 * 3)]/2 + (1 * √(3)/4 + √(3)/6 * 1/2 * 1/3 * 3)
 π/3 – √(3)/6 + 2√(3)/6 = π/3 + √(3)/6.
A semicircle with radius 4 has four isosceles triangles. Each triangle has the base of a fourth of the
radius, and each triangle has the property that its vertex containing the vertex angle is tangent to a
point on the circle. Find the area of the blue region.
Here’s a good problem-solving tactic concerning circles that I like to call “playing with the radius.” We
clearly need to find the area of the triangles we see here and subtract that from the area of the
semicircle. However, we know the bases of each triangle but not the heights. If we toy around with the
radius, we discover some things.
We can drop altitudes for each of these triangles and use the Pythagorean theorem to find the heights.
For the shorter one,
32 + h12 = 42  h1 = √(7).
For the taller one,
12 + h22 = 42  h2 = √(15).
Now, we find the total area of each of the triangles, which is
2(1/2)(2)(√(15)) + 2(1/2)(2)(√(7)) = 2√(7) + 2√(15).
The area of the semicircle is (1/2)(π)(42) = 8π. Finally, the area of the shaded region is
8π – (2√(7) + 2√(15)) = 8π – 2√(7) – 2√(15).
Right triangle ABC has its incircle tangent to points M on AB, N on BC, and P on CA. If AB = 25, BC = 7,
and CA = 24, find the perimeter of triangle MNP.
One thing that will help is to find the inradius of the triangle. The area of triangle ABC is (1/2)(7)(24) =
84, and the semi-perimeter of triangle ABC is 56/2 = 28. Thus, the inradius is 84/28 = 3. We can look at a
diagram to get further in solving this problem.
B
M
N
O
A
P
C
What we see here is that BM ≅ BN, PC ≅ CN, and AM ≅ AP, from the two-tangent theorem. We can use
Pythagoras and Ptolemy repeatedly to get PM and MN. We know we can use Ptolemy because AMOP
and BNOM are cyclic quadrilaterals from opposite angle pairs m∠AMO = m∠APO = 90˚, and m∠BMO =
m∠BNO = 90˚. As for NP, we know that PNCO is a square (if we call O the incenter) because PC ≅ CN,
both are congruent to the radii, and we have 90-degree angles in OPC and ONC. Thus, PN = 3√2.
We now find PM and MN. Starting with PM,
AP = AM = 24 – 3 = 21.
AO = √(212 + 32) = √(450) = 15√2.
By Ptolemy,
(MP)(15√2) = 3(21) + 3(21)  MP = 126/(15√2) = (21√2)/5.
Now we deal with MN.
BM = BN = 7 – 3 = 4.
BO = √(42 + 32) = 5.
Again, by Ptolemy,
(MN)(5) = 3(4) + 3(4) = 24/5.
The perimeter of triangle MNP is 24/5 + (21√2)/5 + (15√2)/5 = [24 + 36√2]/5.
This next problem will bunch together Ptolemy and Power of a Point. What this problem will
involve might be a bit killer, so it’s more suited for the AIME. Beware a large jump in difficulty to this
from the last, and just follow along.
A circle with radius 12 and center O is tangent to AB at point P in triangle ABC. It passes through AC at
points M and C and passes through BC at N and C. Also, m∠ACB = 81˚, and m∠CBA = 39˚. If MC = 12, CN
= 22, AB = 41, and MN = 20, what is the perimeter of triangle BNP?
P
A
M
O
B
N
C
Brace yourselves.
F
A
P
B
M
O
N
C
The diagram looks a bit scary, but we’ll color it so that it’s less scary. The blue will involve Ptolemy since
we see a cyclic quadrilateral from what we colored in blue. The red will involve Power of a Point. Angle
BAC is 180 – 81 – 39 = 60˚. We can safely assume that MO || AP because MOC is equilateral (MC is
congruent to the radius of the circle, so if we connect two radii to make a triangle, we have that MOC is
equilateral), meaning that m∠OMC = 60˚ = m∠BAC. With that aside, let’s get started with our
calculations, particularly with Ptolemy for NP. To find CP, we can use the Pythagorean theorem with
triangle FCP, assuming that F is the foot of the altitude from C to AP. Doing so gets us:
62 + (12 + 6√3)2 = CP2  CP = 12√(2 + √3).
We can get rid of this radical under a radical by using our a2 + 2ab + b2 method. Doing so, we have:
a2 + b2 = 2
2ab = -√3  b = (√3)/2a
2
2
a + 3/4a = 2  4a4 – 8a2 + 3 = 0  (2a2 – 3)(2a2 – 1) = 0.
We want a positive result for CP, so a = (√6)/2 or (√2)/2. CP thus equals 6√6 + 6√2.
Now, by Ptolemy,
(12√2)(22) + (12)(PN) = (20)(6√6 + 6√2)
 12(PN) = 120√6 – 144√2  PN = 10√6 – 12√2.
We then use Power of a Point. We know that PB = 41 – 18 = 23. Power of a point gets us:
232 = BN(BN + 22)  (BN)2 + 22(BN) – 529 = 0
Now, we use the quadratic formula to find BN. That gets us:
[-22 + √(484 + 2116)]/2 = 5√26 – 11.
The perimeter of triangle BNP is 5√26 – 11 + 23 + 10√6 – 12√2 = 5√26 + 10√6 – 12√2 + 12.
Note: You can detect “AMC” when reading the letters on side AC left to right.
TIPS:
 Please remember that both pairs of opposite angles of a quadrilateral have to add up to 180, not
pairs of adjacent angles.
 For finding tangent-tangent, secant-secant, and secant-tangent intersection angles, it’s always the
measure of half the major arc minus the measure of the minor arc. The only case where you have
to add the minor and major arcs together and divide that by two is when you have two secants
intersecting in the circle.
 Play with your radius, kids! Connect radii to different points to see if anything useful happens.
 Watch your math when calculating for semicircles/circle sectors. Make sure you get your
fractions right and remember to distribute your negative signs when subtracting an entire
quantity.
 Sometimes, you might have to square radical expressions, like (2 + √3). It seems obvious, and I
should have said this in our algebra section, but always FOIL. Don’t just square the 2 and the √3
and add them to make 7. You’d hate to make a mistake like that.