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Overview

Triangles are easy to draw and easy to underestimate. On the AMC 8 and early AMC 10, a triangle problem is often not asking for a clever theorem. It is asking whether you can choose the right base, find the matching height, keep track of the perimeter, or notice when three lengths cannot form a triangle at all.

This module covers the triangle facts that show up constantly: area, perimeter, the triangle inequality, and the basic shortcuts for isosceles and equilateral triangles.

Definitions and Key Ideas

A triangle is a polygon with three sides and three angles. If its side lengths are a,b,ca,b,c, then its perimeter is

a+b+c.a+b+c.

The most important area formula is

[ABC]=12⋅base⋅height.[ABC]=\frac{1}{2}\cdot \text{base}\cdot \text{height}.

The height must be perpendicular to the base. It does not have to be drawn inside the triangle, and the base does not have to be the side that looks horizontal.

A B C D base height

For side lengths, the key condition is the triangle inequality:

a+b>c,b+c>a,c+a>b.a+b>c,\qquad b+c>a,\qquad c+a>b.

Equivalently, if two sides have lengths aa and bb, then the third side cc must satisfy

∣a−b∣<c<a+b.|a-b|<c<a+b.

The inequalities are strict. If the longest side equals the sum of the other two, the triangle has flattened into a line segment, so it is not a triangle.

Area: Choosing the Right Base

Any side of a triangle can be used as the base. Once the base is chosen, the height is the perpendicular distance from the opposite vertex to the line containing that base.

For example, if AB=18AB=18 and the altitude from CC to ABAB is 77, then

[ABC]=12⋅18⋅7=63.[ABC]=\frac{1}{2}\cdot 18\cdot 7=63.

A common contest trick is to include extra side lengths or angle information. Use the information the problem gives you immediately.

Triangle Inequality

Suppose two sides are 66 and 1010. The third side xx must satisfy

∣10−6∣<x<10+6,|10-6|<x<10+6,

so

4<x<16.4<x<16.

If xx must be an integer, the possible values are

5,6,7,8,9,10,11,12,13,14,15.5,6,7,8,9,10,11,12,13,14,15.

There are 1111 such values.

A B C longest side one side other side

The intuition is rather simple: the two shorter sides have to be long enough to actually meet above the longest side.

Isosceles and Equilateral Triangles

An isosceles triangle has two equal sides. The angles opposite those equal sides are also equal.

B C A base angle base angle

So if AB=ACAB=AC, then

∠B=∠C.\angle B=\angle C.

An equilateral triangle has all three sides equal and all three angles equal to 60∘60^\circ. Dropping an altitude in an equilateral triangle splits it into two congruent 3030-6060-9090 triangles.

A useful isosceles shortcut: in an isosceles triangle, the altitude from the vertex angle to the base also bisects the base. For instance, in a triangle with sides 10,10,1210,10,12, the altitude to the base splits the base into 66 and 66. Then the height is

102−62=8,\sqrt{10^2-6^2}=8,

so the area is

12⋅12⋅8=48.\frac{1}{2}\cdot 12\cdot 8=48.

Perimeter and Hidden Equations

Perimeter problems often become one-line equations. If an isosceles triangle has equal sides x,xx,x and base 1010, and its perimeter is 3838, then

2x+10=38,2x+10=38,

so

x=14.x=14.

After solving, it is still worth checking the triangle inequality. Here 14+10>1414+10>14 and 14+14>1014+14>10, so the triangle is valid.

Worked Example

Two sides of a triangle have lengths 55 and 88. What are the possible integer lengths for the third side?

By the triangle inequality,

∣8−5∣<x<8+5,|8-5|<x<8+5,

so

3<x<13.3<x<13.

The integer values are

4,5,6,7,8,9,10,11,12,4,5,6,7,8,9,10,11,12,

so there are

9\boxed{9}

possible lengths.

More Examples

Example 1: Area Without Extra Work

Triangle ABCABC has AB=20AB=20. The perpendicular distance from CC to line ABAB is 99. Find [ABC][ABC].

The area is

12⋅20⋅9=90.\frac{1}{2}\cdot 20\cdot 9=\boxed{90}.

We do not need to know where the altitude lands on the line.

Example 2: Isosceles Angle Chase

In triangle ABCABC, suppose AB=ACAB=AC and ∠A=44∘\angle A=44^\circ. Find ∠B\angle B.

Since AB=ACAB=AC, the base angles are equal, so ∠B=∠C\angle B=\angle C. Thus

∠B+∠C=180∘−44∘=136∘,\angle B+\angle C=180^\circ-44^\circ=136^\circ,

and

∠B=68∘.\angle B=\boxed{68^\circ}.

Example 3: Checking Whether a Triangle Exists

Can a triangle have side lengths 4,7,124,7,12?

No. The two shorter sides have sum

4+7=11,4+7=11,

but 11<1211<12. The two shorter sides cannot reach each other, so no such triangle exists.

Strategies

  • If the problem asks for area, look for a base-height pair before doing anything else.
  • If a side length is unknown, write the triangle inequality range.
  • If two sides are equal, mark the opposite angles equal.
  • If two angles are equal, mark the opposite sides equal.
  • In an equilateral triangle, immediately write every angle as 60∘60^\circ.
  • In an isosceles triangle, consider dropping the altitude to the base; it often splits the base in half.

Warnings

  • Using a slanted side as a height. Heights must be perpendicular.
  • Forgetting that triangle inequalities are strict.
  • Checking only one triangle inequality. Usually the only dangerous one is whether the two shorter sides are longer than the longest side, but make sure you have identified the longest side correctly.
  • Assuming an isosceles triangle is equilateral. Equal sides give equal base angles, not necessarily 60∘60^\circ angles.
  • Solving a perimeter equation but forgetting to check that the side lengths can actually form a triangle.

Practice Problems

StatusSourceProblem NameDifficultyTags
AMC 8Easy
Show TagsArea, Triangles
AMC 8Easy
Show TagsArea, Coordinate geometry, Triangles
AMC 8Medium
Show TagsArea, Coordinate geometry, Triangles
AMC 8Medium
Show TagsArea, Isosceles triangles, Triangles
AMC 8Medium
Show TagsIsosceles triangles, Triangle inequality, Triangles
AMC 8Medium
Show TagsArea, Equilateral triangles, Triangles
AMC 10Medium
Show TagsPerimeter, Triangle inequality, Triangles
AMC 10Medium
Show TagsIsosceles triangles, Triangle inequality, Triangles

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