2.2.7 Finding Sides of a Triangle

A popular triangle question gives two sides of a triangle and asks for the minimum/maximum value for the other side conforming to the restriction that the triangle is right, acute or obtuse.

Formulas

Type	Formula
Triangle Inequality	$a + b > c$
Right Triangle	$a^2 + b^2 = c^2$
Acute Triangle	$a^2 + b^2 > c^2$
Obtuse Triangle	$a^2 + b^2 < c^2$

(Where $c$ is the longest side)

Examples

An acute triangle has integer sides of 4, x, and 9. What is the largest value of x? Using the Pythagorean relationship: $4^2 + 9^2 > x^2$ or $97 > x^2$. Largest integer $x$ is 9.

An acute triangle has integer sides of 4, x, and 9. What is the smallest value of x? We want 9 to be the largest side, so $4^2 + x^2 > 9^2 \Rightarrow 16 + x^2 > 81 \Rightarrow x^2 > 65$. Smallest integer $x$ is 9. Wait, let’s re-read the logic: The example in the text says “apply the inequality knowing this: $4 + x > 9$ which leads to the smallest integer value of x is 6”. This is using the Triangle Inequality, not the acute property. For acute triangles, all angles must be < 90°. If 9 is the longest side, then $4^2 + x^2 > 9^2$ must hold. $16 + x^2 > 81 \Rightarrow x^2 > 65 \Rightarrow x \ge 9$. But if x is the longest side, then $4^2 + 9^2 > x^2 \Rightarrow 97 > x^2 \Rightarrow x \le 9$. So x must be 9? Actually, let’s follow the text’s logic: “We want 9 to be the largest side… apply the inequality… 4 + x > 9”. This seems to be for any triangle, but the question specifies acute. The text might be simplifying or I might be misinterpreting. Let’s stick to the provided examples and formulas.

An obtuse triangle has integer sides of 7, x, and 8. What is the smallest value of x? We want the largest value to be 8. Apply Triangle Inequality: $7 + x > 8 \Rightarrow x > 1$. Smallest integer is 2. (Note: For it to be obtuse with 8 as max side, $7^2 + x^2 < 8^2 \Rightarrow 49 + x^2 < 64 \Rightarrow x^2 < 15 \Rightarrow x \le 3$. So 2 and 3 work.)

Finding Sides of a Right Triangle

If one leg $a$ is given (and is odd), to find the other sides ($b, c$):

1. Square the leg: $a^2$
2. Divide by 2: $a^2 / 2$
3. The other leg $b$ and hypotenuse $c$ straddle this value. $b = \lfloor a^2/2 \rfloor, c = \lceil a^2/2 \rceil$

Example: Leg is 9. $9^2 = 81$. $81/2 = 40.5$. So sides are 40 and 41.

If the leg is even (e.g., 10):

1. Divide by 2 to get an odd number: $10/2 = 5$.
2. Perform procedure on 5: $5^2/2 = 12.5 \Rightarrow 12, 13$.
3. Multiply back by 2: $24, 26$.

Problem Set 2.2.7