VMHS Math Circle
XI. The Grand Geometry Review (Part I: Synthetic Geometry-Lines and Triangles)
When people refer to synthetic geometry, they’re talking about the geometry that doesn’t have to
do with the coordinate plane. Geometry will be a bit of a bulky review because AMC problems dealing
with geometry tend to tie together many different aspects of geometry. While quite a bit of this may seem
like review, try to bear with me because there will be some problem-solving topics that will serve as
important for the AMC.
Let’s start off with something simple.
Rays, Lines, and Transversals
This is basic geometry knowledge. We all know what a point is: a specific location in space
that we decide to mark. Now, define a line as a straight one-dimensional figure with no thickness
that extends infinitely in both directions. If a line has two endpoints, we call it a line segment. If a
line has one endpoint, we call it a ray.
With the basics aside, there are some notable things about lines that you should keep in mind.
Transversals with Two Parallel Lines:
A transversal is when one line passes through two other lines. If those two other lines are parallel, then
remember the following. Note that parallel lines are denoted by a “>.” We use “∠” to denote an angle.
7 8
5 6
3 4
1 2
The “congruent to” sign is ≅.
Alternate exterior angles: congruence between any pair of angles that are on opposite sides of the
transversal but are outside of the two parallel lines; ∠1 ≅ ∠8, ∠2 ≅ ∠7 from this.
Alternate interior angles: congruence between any pair of angles that are on opposite sides of the
transversal but are inside the two parallel lines; ∠3 ≅ ∠6, ∠4 ≅ ∠5 from this.
Consecutive interior angles: the measures of two interior angles that are on the same side of the
transversal add up to 180 degrees; m∠3 + m∠5 = 180, m∠4 + m∠6 = 180 from this.
It is crucial to know the following for angles formed by the intersection of two lines.
4 1
3 2
Linear Pair: if two angles form a line, the measures of those two angles add up to 180 degrees; m∠1 +
m∠2 = 180, m∠2 + m∠3 = 180, m∠3 + m∠4 = 180, m∠4 + m∠1 = 180 from this.
Vertical Angles: opposite angles caused by the intersection of two lines are congruent; ∠1 ≅ ∠3, ∠2 ≅
∠4 from this.
Perpendicular Lines: if an angle formed by the intersection of two lines is 90 degrees, all other angles
are 90 degrees. This applies to transversals as well. If m∠1 = 90, m∠2 = 90, m∠3 = 90, and m∠4 = 90.
Lines and transversals form a very simple and compact topic. As such, AMC problems dealing
solely with lines in synthetic geometry are scarce. Let’s move on to something more relevant to the AMC.
Triangles
Loads of AMC geometry problems deal with triangles. Because of this, there are a lot of things
and properties that you should know about triangles. Let’s deal with the less obscure ones first.
Triangle Sum Theorem:
Simply enough, the measures of three interior angles of a triangle add up to 180 degrees.
Triangle Inequality:
Given a triangle with side lengths a, b, and c, with “a” and “b” being any two sides, it must be true that a
+ b < c, as well as any rearrangement of this inequality. If a + b ≥ c, then the triangle becomes a line.
Basic Perimeter/Area of a Triangle:
Perimeter = a + b + c, given sides a, b, and c.
Area = bh/2, given a base “b” and a height “h.”
Pythagorean Theorem:
Given a right triangle with legs “a” and “b” with a hypotenuse (longest side) “c,”
a2 + b2 = c2.
Here’s some more obscure stuff.
Angle Bisector Theorem:
Given a triangle with a specific angle bisector,
c
a
d
b
a/b = c/d.
Thales’s Theorem:
If a right triangle is inscribed in a circle, the hypotenuse is always that circle’s diameter.
AMC 12 Only
Other Triangle Area Formulas:
A = √(s(s – a)(s – b)(s – c)), given side lengths a, b, and c and that “s” is half the triangle’s perimeter
A = [bc(sin(A))]/2 given any angle “A” and any adjacent side lengths “b” and “c”
A = (abc)/(4R), given side lengths a, b, and c, and circumradius R
A = rs, given inradius “r” and that “s” is half the perimeter
Please remember the first two formulas! The last two are more obscure and are more abundant in the
AIME.
Trig for Triangles:
Warning: I know what’s coming up is going to be a lot, but please know that it’s for the best. If you’ve
learned pre-calculus and hated trig identities, I am legitimately sorry. However, the AMC won’t be,
especially for AMC 12 takers. Please remember these formulas!
A
b
c
C
B
a
Use this triangle for reference.
Law of Sines: sinA/a = sinB/b = sinC/c, given angles A, B, and C and opposite sides a, b, and c.
Law of Cosines: a2 = b2 + c2 – 2bc(cosA).
Note that for the Law of Sines, watch “ambiguous cases.” These don’t pop up in the AMC very much
since trig is used in a problem solving context here, and using the Law of Sines is uncommon. Google
“ambiguous case” and learn it for yourself if you don’t understand what I’m talking about.
Half Angle Formulas:
sin(a/2) = √[(1 – cos(a))/2]
cos(a/2) = √[(1 + cos(a))/2]
tan(a/2) = (1 – cos(a))/sin(a)
Double Angle Formulas:
sin(2a) = 2sin(a)cos(b)
cos(2a) = cos2(a) – sin2(a)
tan(2a) = 2tan(a)/(1 – tan2(a))
Addition/Subtraction:
sin(a + b) = sin(a)cos(b) + cos(a)sin(b) | sin(a – b) = sin(a)cos(b) – cos(a)sin(b)
cos(a + b) = cos(a)cos(b) – sin(a)sin(b) | cos(a – b) = cos(a)cos(b) + sin(a)sin(b)
tan(a + b) = (tan(a) + tan(b))/(1 – tan(a)tan(b)) | tan(a – b) = (tan(a) – tan(b))/(1 + tan(a)tan(b))
There are also various angle/side congruence/similarity theorems you should be familiar with.
Note that congruence is when two figures are identical in shape and size to each other (though you
might need to rotate/turn one figure to make it match the other). If two polygons “R” and “S,” both with
“n” sides, are congruent, and we line up corresponding parts, R1 = S1, R2 = S2, R3 = S3, … , Rn = Sn, in
which our sides are labeled 1, 2, 3, … , n.
Two figures are similar when you can scale down or scale up one figure (and possibly turn it)
to make it identical to the other. If two polygons “R” and “S,” both with “n” sides, are similar, and we
line up corresponding parts, R1/S1 = R2/S2 = R3/S3 = … = Rn/Sn, in which our sides are labeled 1, 2, 3, … ,
n.
To explain these theorems, I’ll use pictures. Keep the definitions of similarity and congruence in
mind.
SAS: Congruence is established when the corresponding sides are congruent, as well as the angle between
them. Similarity is established when the pairs of corresponding sides are in the same ratio.
ASA (AA for similarity): Congruence is established when the corresponding angles are congruent, as
well as the side between them. Similarity is established when these conditions are met, but the
corresponding sides are not necessarily equal.
SSS: Congruence is established when the corresponding sides are congruent. Similarity is established
when the corresponding sides are in the same ratio.
AAS (AA for similarity): Congruence is established when the corresponding angles are congruent, as
well as the side outside of them. Similarity is established when these conditions are met, but the
corresponding sides are not necessarily equal.
HL: Congruence is established when corresponding hypotenuses and legs are congruent. Similarity is
established when corresponding hypotenuses and legs are in the same ratio.
Know how angles are related to sides. That is, know that if two corresponding angles are
congruent, then corresponding sides are congruent or similar. Keep in mind that the larger an
angle gets, the longer its corresponding side. Also, know the properties of isosceles and equilateral
triangles. Know that if “n” angles form a line, the measures of those angles add up to 180!
Here are some definitions you must know also.
Special Cevians/Lines:
Perpendicular Bisector: A line going through a side’s midpoint that is perpendicular to that side.
Angle Bisector: A line from a vertex of an angle that splits the angle in two halves.
Median: A line from a vertex of a triangle that passes through the midpoint of that angle’s opposite side
Altitude: A line from the vertex of a triangle that is perpendicular to that angle’s opposite side
“Centers”:
Incenter: Where the angle bisectors intersect
Circumcenter: Where the perpendicular bisectors intersect (useful for mass points), center of the
circumcircle of a triangle
There was a document earlier about something called mass points. The reason why it was an
entire document rather than a section here was that it took a while to explain, seeing that it was a
completely different tactic. As I said in that document, it really helps with triangle ratios and areas. Since
this is basically all you need to know for triangles, let’s take on some practice problems. Scratch “some.”
Let’s have quite a few sample problems in here to help you get used to AMC geometry. What I went over
was pretty simply stated. However, the AMC can take this very, very far.
Expect earlier questions to be SAT-type questions.
Consider two triangles: ABC and CBD. Given that AB = AC = 4, BC = CD = 8, BD = 16, and m∠ABC = 20˚,
what is m∠ACD in degrees?
Always draw diagrams. They’re very important in geometry. First thing:
B
20˚ 20˚
4
A 140˚
4
16
8
20˚ 140˚
20˚
C
D
8
Triangles ABC and BCD are similar. Both are isosceles triangles with the ratios of two sides of two
different lengths being 1:2. Thus, their angles are congruent, so m∠ACD = m∠ACB + m∠BCD = 20 + 140 =
160.
Right triangle ABC has BC = 4. If the altitude of triangle ACD to AC is congruent to AB and ACD has an
area of 3, what is the measure of AC?
Diagram time!
D
C
4
x
A
y
x
B
Okay, first thing we do is set up a system of equations. Get into the gear of being ready to solve
systems of equations in geometry problems. You need to recognize that you can do this when you can
establish two different relationships with only two unknowns.
(1/2)(xy) = 3  y = 6/x
x2 + 4 2 = y 2
We substitute:
x2 – (6/x)2 + 16 = 0  x4 + 16x2 – 36 = 0.
Solving gets us (x2 + 18)(x2 – 2), so our only positive root is x = AB = √(2). Thus, AC = y = 6/√(2) = 3√(2).
Later questions are going to be taken further.
Right triangle ABC has m∠ABC = 90˚. An altitude is dropped from B to AC, with its foot being D. An
altitude is then dropped from D to BC with its foot being E. Three more altitudes are dropped: one from
E to BD with foot F, one from F to BE with foot G, and one from F to DE with foot H. AB = 4, and BC = 3.
The area of quadrilateral FGEH can be expressed as (abcd)/ef in simplest form with a, b, c, d, and e being
positive integers and a, c, and e being as low as possible. Find a + b + c + d + e + f.
There are a couple things to note about diagrams. If you’re having trouble reading one, enlarge it.
Also, try drawing a diagram to scale using your ruler and/or protractor (It makes things clearer, and
you never know, you may cheat your way toward the answer using a simple pencil and ruler). Don’t
take too long; you know that 75 minutes is not a lot of time to do 25 problems.
B
G
E
F
H
A
C
D
Because we’re dropping a bunch of altitudes here, it’s nice to have big enough diagrams to work with.
One thing good to find is that FGEH is a rectangle because angles FGE, GEH, and EHF are formed from
altitudes. If a quadrilateral has three 90˚ angles, the fourth angle is 360 – 270 = 90˚. Now, we perform
calculations using the fact that all the triangles we made are similar to triangle ABC (this is because
triangle DBC had a right angle and angle C in common with triangle ABC, meaning AA similarity, so we
can apply this analogously with all other triangles we make by dropping altitudes). AC = 5 because ABC is
a 3-4-5 right triangle.
CD/3 = 3/5  CD = 9/5; BD = 12/5
DE = 36/25; HF = 432/625, EH = 576/625.
Here, I use the fact that the altitude to the hypotenuse of a 3-4-5 right triangle has length 12c/25, with
“c” being the hypotenuse. Since FGEH is a rectangle, all we need to do is multiply the lengths of HF and
EH. Doing so any simplifying gets us: 21035/58. Thus, the answer is 2 + 10 + 3 + 5 + 5 + 8 = 33.
Triangle ABC has points D and E on ABC such that angle DAB is a right angle, and AE bisects angle CAB. If
AB = 30, and AD = 40, and DE = EB, what is CE?
A
C
D
E
F
B
If ED = EB, then we can say that E is the midpoint of DB. By Thales’s theorem, we can inscribe ABD in a
circle, with DB being a diameter. From this, E is the center of this circle, so EA = ED = EB. DB is 50
because triangle ADB is a 3-4-5 right triangle, so DE = EB = EA = 25. The altitude of ADB from A to BD
(let’s call the foot of the altitude “F”) is (1/2)(50)(AF) = 600  50(AF) = 1200  AF = 24. Hence, AEF is a
7-24-25 triangle, so EF = 7. Now, we set up a system of equations using Pythagoras and the angle
bisector theorem.
(7 + CE)2 + 242 = AC2
AC/CE = 30/25 = 6/5  AC = (6/5)(CE)
We then substitute and perform some not-too-pretty looking calculations:
(7 + CE)2 + 576 = (36/25)(CE2)
 49 + 14(CE) + (CE)2 + 576 = (36/25)(CE2)
 (11/25)(CE2) – 14(CE) – 625 = 0
Time for some quadratic formula. Our positive answer is:
(14 + √(196 + 1100))/(22/25) = (14 + √(1296))/(22/25) = 50/(22/25) = 625/11.
However, there are some problems that require some insane thinking. I call this “geometric
manipulation,” similar to algebraic manipulation in that it’s just extraordinary. However, there are some
situations in which it is clear that you must reflect, dilate, translate, cut, or rotate whatever shape you’re
given: some problems in which you are asked to find the minimum distance and some problems that
look wacky in a way that nothing you try to do works. Let’s bring in some examples. I’d like to start
off with one problem that basically originated about 2000 years ago: Heron’s Shortest Path problem.
Points A and B both lie not on line L, but on one side of horizontal line L. Given that C lies on L, AB = 12,
the distance from B to L is 8, and the distance from A to L is 4, what is the minimum value of AC + BC?
This is what you have to do.
B
C
A
L
C
A’
D
That’s right. You have to reflect A (or B) across line L and choose a C such that A, C, and B are collinear.
It’s easy to imagine that a straight line with A, B, and C has a shorter value of AC + BC than a V-shape. All
abstractness aside, we apply Pythagoras. If we extend A’ horizontally and B vertically and call the
intersection D, we know that BD = 4 + 4 + (8 – 4) = 12. If we extend A horizontally and B vertically and
call the intersection P, we know that AP = √(122 – 42) = √(144 – 16) = √(128) = 8√2.
Using Pythagoras again, A’C + BC = A’B = √((8√2)2 + 122) = √(128 + 144) = √(272) = 4√(17).
Others are even crazier.
Isosceles right triangle ABC has a right angle at A. Line L passes through B in a way that makes BC ⊥ L.
There are points D inside ABC and E outside ABC that exist such that line DE is parallel to line L, and BC
bisects segment DE. Also, there is a point F on line L, closer in distance to point E than to point D, such
that BF = BD. Given that BD = 4, and EF = 6, find DF.
Well, if we have triangles and perpendicular lines, it’d help to make BC vertical.
F
B
A
L
D
E
C
Now, this looks a bit wacky and disjointed. We don’t know any angle measures besides 45, 45, and 90,
so we can’t necessarily use trig (if anyone reading this finds a solution with trig, a dear congratulations
to him or her). We can’t use Pythagoras either, since we don’t know a lot of sides. However, I’ll show
you what to do. This is why “geometric manipulation” is at least as good as algebraic manipulation.
Behold!
F
B
A
D
L
E
C
“Whoa” is right. What we do is reflect and move the triangle in all these ways. The images of points D, E,
and F all have the same distance to B from the original D, E, and F from our transformations. Also, from
this, EF is congruent to the segment that joins D and the reflected image of F. Hence, we can safely draw
a circle around all of these points. By Thales, a right triangle is formed from the image of F, as well as
points D and F. Now we can use Pythagoras. Doing so, we have:
62 + (DF)2 = 82  36 + (DF)2 = 64  (DF)2 = 28  DF = 2√7.
For the AMC 12, geometry problems might involve trigonometry. This next problem will be a bit
overkill for the AMC computation-wise, but it bunches a lot of needed trig skills together. Bear with me,
because this problem is arguably crazier than the last. This one I won’t label “AMC 12 Only” because
there is an AMC 10 trick that takers need to know about right triangles. I’ll highlight this in blue.
Triangle ABC has m∠C = 45˚. A 30-60-90 triangle EFG is embedded within ABC such that BC contains FG,
EF ⊥ BC, E lies on AB, and m<FEG = 60˚. Given that EG = CG = 4, and BC = 10, the side AC can be
expressed in the form [a√b + c√d]/e, all in simplest terms. Find a + b + c + d + e.
We start with this:
A
E
4
C
45˚
60˚
30˚
B
F
G
4
10
Again, this looks a bit abstract. We know that we can’t magically get the length of AC from what we have
here. Let’s do something cool, then, and size this diagram up because we’ll be doing some angle chasing.
A
J
H
135˚ 30˚ 150˚
C
45˚
15˚
15˚
4
4
15˚
15˚
15˚
K
4
E
4
150˚
60˚
30˚
G
F
B
10
All right! First up, we can move EF a ways left to make HK, where H lies on AC. Since EFG is a 30-60-90
triangle, EF = 2. Thus, we get a nice 45-45-90 triangle with HC = 2√2. Now, the hard part is finding AH.
This is where we have to draw all these lines and fish for all these angles. First, we can draw a line from
C to E in order to make a 15-75-90 triangle.
The special thing about 15-75-90 triangles is that you can basically draw a 30-60-90 triangle in it, and get
an isosceles 15-15-150 triangle. This means that if you see a 15-75-90 triangle, draw a 30-60-90 triangle
with the 75 degree angle containing your new 60 degree angle. This way, you get the isosceles 15-15150 triangle that helps simplify any problem you get that has these types of triangles.
Going back to our problem, we can then draw EH parallel to BC to make another 15 degree angle in HEC.
This is because angle CEF is 75 degrees like we found, and we’ve just made a right angle because EH ⊥
EF from EH || BC and EF ⊥ BC. Another nice thing we can do is reflect triangle CEG to make a kite CJEG.
Hence, CJ = JE = 4, and angle ACB is trisected by CJ and CE. Now, we break out some trig to find HJ. We
use a combination of the Law of Sines and sine subtraction.
sin(30˚)/2√2 = sin(45˚ – 30˚)/(HJ)  1/4√2 = [(√2/2)( √3/2) – (√2/2)(1/2)]/(HJ)
HJ = (4√2)[(√2/2)( √3/2) – (√2/2)(1/2)] = 2√3 – 2.
Hence, HE = HJ + JE = 2√3 – 2 + 4 = 2√3 + 2.
Then, we notice that triangles AEH and ABC are similar from EH || BC, ∠AHE ≅ ∠ACB, and ∠AEH ≅
∠ABC. This means that AH/AC = (2√3 + 2)/10. Since we know that HC = 2√2, we can solve for AH and add
HC to it to finally get AC! Doing so gets us:
AH/(AH + 2√2) = (2√3 + 2)/10  AH = [(2√3)(AH) + 2(AH) + 4√6 + 4√2]/10
 (AH)(8 – 2√3) = (4√6 + 4√2)  AH = (32√6 + 32√2 + 24√2 + 8√6)/52 = (10√6 + 14√2)/13.
Adding 2√2 to this gets us AC = (10√6 + 40√2)/13.
Our answer is 10 + 6 + 40 + 2 + 13 = 71.
Note: you could have used Law of Sines for triangle HEC, as well. I just made a kite since it was the first
thing I thought of doing at the time. Math encourages unique, creative solutions anyway!
TIPS:
 I’ve basically said what I wanted throughout this document, but I do want to reiterate some
things. Remember to look for similar triangles, as they are a huge part of AMC geometry
dealing with triangles.
 The Pythagorean Theorem is also popular, but the AMC will twist it in that you might have to
solve a system of equations using it or maybe break out some trig.
 I reiterate: if nothing you do works, remember geometric manipulation! Try some translations or
reflections. Draw new lines, or perform rotations!
 To those who don’t understand my explanation of splitting a 15-75-90 triangle, let’s say I have
this triangle PQR. If angle Q is 75 degrees and angle P is right, I can draw a segment from Q to
PR, call the foot X, and make the measure of angle XQP 60 degrees.
R
X
P
Q