1. **Stating the problem:** We have a sequence of right triangles where the shortest side $a=1$ is fixed. The triangles are: - Triangle 1: sides $(a, b, c) = (1, \sqrt{2}, 1)$ - Triangle 2: sides $(1, \sqrt{3}, \sqrt{2})$ - Triangle 3: sides $(1, \sqrt{4}, \sqrt{3})$ We want to describe the pattern, predict term 4, derive formulas for $b$ and $c$ in terms of the term number $t$, and predict term 2022. 2. **Observing the pattern:** - The shortest side $a=1$ is constant. - The side $b$ in term $t$ appears to be $b_t = \sqrt{t+1}$. - The hypotenuse $c$ in term $t$ appears to be $c_t = \sqrt{t}$. 3. **Verifying the pattern with given terms:** - For $t=1$: $b_1 = \sqrt{2}$, $c_1 = \sqrt{1} = 1$ matches Triangle 1. - For $t=2$: $b_2 = \sqrt{3}$, $c_2 = \sqrt{2}$ matches Triangle 2. - For $t=3$: $b_3 = \sqrt{4} = 2$, $c_3 = \sqrt{3}$ matches Triangle 3. 4. **Predicting Term 4:** - $b_4 = \sqrt{5}$ - $c_4 = \sqrt{4} = 2$ 5. **Deriving the rules:** - Since $a=1$ is constant, - $b_t = \sqrt{t+1}$ - $c_t = \sqrt{t}$ 6. **Using the rules to predict Term 2022:** - $b_{2022} = \sqrt{2023}$ - $c_{2022} = \sqrt{2022}$ 7. **Verification that these form right triangles:** - By Pythagoras theorem, $a^2 + c_t^2 = b_t^2$ - Substitute: $1^2 + (\sqrt{t})^2 = (\sqrt{t+1})^2$ - Simplify: $1 + t = t + 1$ which is true. **Final answers:** - General formulas: $a=1$, $b_t=\sqrt{t+1}$, $c_t=\sqrt{t}$ - Term 4: $(1, \sqrt{5}, 2)$ - Term 2022: $(1, \sqrt{2023}, \sqrt{2022})$