Overview

Right triangles are the absolute backbone of competition geometry. Because one angle is locked exactly at $90^\circ$, right triangles possess incredibly rigid and predictable structural properties. They are the keys to finding distances, calculating areas, and decomposing complex polygons into solvable pieces.

Mastering right triangles means moving beyond just knowing the formula; you must learn to instantly recognize patterns (like triples and special angles) to save massive amounts of time on the AMC.

1. The Pythagorean Theorem

The most famous theorem in mathematics describes the fundamental relationship between the three sides of a right triangle.

If a right triangle has two legs of lengths $a$ and $b$, and a hypotenuse (the longest side, directly opposite the right angle) of length $c$, then:

Geometric Meaning: If you physically draw a square originating from each of the two legs, the total area of those two squares will exactly equal the area of a massive square drawn from the hypotenuse.

2. Pythagorean Triples (The Speed Cheat Codes)

While the Pythagorean theorem works for any real numbers, AMC problem writers love to use integers. A Pythagorean Triple is a set of three positive integers $(a, b, c)$ that perfectly satisfy $a^2 + b^2 = c^2$.

Memorizing the most common primitive triples (triples where the numbers share no common factors) is mandatory for speed. If you recognize two sides of a triple, you can instantly fill in the third without doing any algebra.

The Core Four Primitives:

1. 3 - 4 - 5 (The most common)
2. 5 - 12 - 13
3. 8 - 15 - 17
4. 7 - 24 - 25

Scaled Triples:

If you multiply a triple by any integer constant $k$, the resulting numbers also form a right triangle. For example, multiplying the 3-4-5 triangle by $2$ gives the 6-8-10 triangle. Multiplying it by $10$ gives the 30-40-50 triangle.

Pro-Tip: If you see a hypotenuse of $51$ and a leg of $24$, do not calculate $51^2 - 24^2$. Notice that $51 = 17 \times 3$ and $24 = 8 \times 3$. This is an 8-15-17 triangle scaled by 3! The missing leg is simply $15 \times 3 = 45$.

3. Special Right Triangles

When the angles of a right triangle are specifically chosen, the side lengths lock into fixed, predictable ratios involving square roots. These appear in almost every competition geometry problem involving hexagons, equilateral triangles, or squares.

The 45-45-90 Triangle (Isosceles Right)

Created by cutting a square in half along its diagonal. Because the two base angles are equal ($45^\circ$), the two legs are equal.

• Ratio: $x : x : x\sqrt{2}$
• Rule: To get the hypotenuse, multiply the leg by $\sqrt{2}$. To get the leg, divide the hypotenuse by $\sqrt{2}$.

The 30-60-90 Triangle

Created by cutting an equilateral triangle exactly in half down its altitude.

• Ratio: $x : x\sqrt{3} : 2x$
• Rule: The short leg ($x$) is always opposite the $30^\circ$ angle. The hypotenuse is exactly double the short leg. The long leg (opposite the $60^\circ$ angle) is the short leg multiplied by $\sqrt{3}$.

Worked Examples

Example 1: Cascading Triangles

Problem: In the figure above, $\angle B = 90^\circ$ and $\angle ACD = 90^\circ$. If $AB = 4$, $BC = 3$, and $CD = 12$, what is the length of $AD$?

Solution: We must solve this in two steps. First, look at $\triangle ABC$. It has legs $3$ and $4$. Recognizing the primitive 3-4-5 Pythagorean triple, we instantly know the hypotenuse $AC = 5$.

Now, move to $\triangle ACD$. It is a right triangle with legs $AC = 5$ and $CD = 12$. Recognizing another core primitive triple, 5-12-13, we immediately deduce that the hypotenuse $AD$ must be $13$.

Example 2: Altitudes of Equilateral Triangles

Problem: An equilateral triangle has a side length of $10$. What is its area?

Solution: To find the area, we need the height (altitude). If we drop an altitude from the top vertex of an equilateral triangle, it splits the base perfectly in half and bisects the $60^\circ$ top angle into two $30^\circ$ angles.

This creates two 30-60-90 right triangles. The hypotenuse is the original side length: $10$. The short leg (half the base) is: $5$. Using our 30-60-90 rule ($x : x\sqrt{3} : 2x$), the long leg (the altitude) is the short leg times $\sqrt{3}$, which is $5\sqrt{3}$.

Now, calculate the area of the full equilateral triangle:

Example 3: Finding Missing Legs (The Subtraction Trap)

Problem: A right triangle has a hypotenuse of $17$ and one leg of length $9$. What is the length of the other leg?

Solution: A common mistake here is to blindly add the squares ($17^2 + 9^2$), forgetting that $17$ is the hypotenuse ($c$). The correct setup is:

$$a^2 = 289 - 81$$

$$a^2 = 208$$

$$a = \sqrt{208}$$

To simplify $\sqrt{208}$, look for perfect square factors. $208 / 16 = 13$.

$$a = \sqrt{16 \cdot 13} = 4\sqrt{13}$$

(Notice: This was NOT an 8-15-17 triple! The hypotenuse was 17, but the leg was 9, not 8 or 15. Always verify the numbers fit the triple exactly before assuming).

Common Pitfalls

• Adding instead of subtracting: If you are given the hypotenuse and asked for a leg, you must subtract the square of the known leg from the square of the hypotenuse ($a^2 = c^2 - b^2$).
• Assuming visual right angles: Just because a triangle looks like a right triangle in a diagram does not mean it is one. You cannot use $a^2 + b^2 = c^2$ unless the problem explicitly states there is a $90^\circ$ angle, or if you can prove it via other geometry rules.
• Applying triples to the wrong sides: For a 3-4-5 triangle, the $5$ must be the hypotenuse. If a right triangle has legs of $3$ and $5$, the hypotenuse is $\sqrt{3^2 + 5^2} = \sqrt{34}$, not $4$.

Practice Problems