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.
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.
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,
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!
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:
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!
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 à y = 6/x
x2 + 42 = y2
We substitute:
x2 – (6/x)2 + 16 = 0 à 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.
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 à 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?
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 à 50(AF) = 1200 à 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 à AC = (6/5)(CE)
We then substitute and perform some not-too-pretty looking calculations:
(7 + CE)2 + 576 = (36/25)(CE2)
à 49 + 14(CE) + (CE)2 + 576 = (36/25)(CE2)