Problem 1451

1. Given Information and Setup:
- We have a square $ABCD$ with an interior point $P$.
- The distances from $P$ to the vertices are given as $PA = a$, $PB = b$, and $PC = b + c$.
- The relationship $a^2 = b^2 + c^2$ holds.

2. Objective:
- Prove that the angle $\angle BPC$ is a right angle.

3. Constructing the Triangle:
- Consider the right and isosceles triangle $PBE$ where $E$ is a point such that $\angle BPE = 45^\circ$ and $PE = PB = b$.

4. **Using the Law of Cosines in $\triangle PEC$:**
- We need to find the length $PE$ and use the law of cosines in $\triangle PEC$.
- Since $PE = PB = b$ and $\angle BPE = 45^\circ$, we can use the law of cosines in $\triangle PEC$.

5. Applying the Law of Cosines:
- In $\triangle PEC$, we have:
$$PC^2 = PE^2 + EC^2 - 2 \cdot PE \cdot EC \cdot \cos(\angle PEC)$$
- Substituting the known values:
$$(b + c)^2 = b^2 + a^2 - 2 \cdot b \cdot a \cdot \cos(\angle PEC)$$
- Given $a^2 = b^2 + c^2$, we substitute $a^2$:
$$(b + c)^2 = b^2 + (b^2 + c^2) - 2 \cdot b \cdot a \cdot \cos(\angle PEC)$$
$$(b + c)^2 = 2b^2 + c^2 - 2 \cdot b \cdot a \cdot \cos(\angle PEC)$$

6. Simplifying the Equation:
- Expanding $(b + c)^2$:
$$b^2 + 2bc + c^2 = 2b^2 + c^2 - 2 \cdot b \cdot a \cdot \cos(\angle PEC)$$
- Canceling out $b^2 + c^2$ from both sides:
$$2bc = 2b \cdot a \cdot \cos(\angle PEC)$$
- Dividing both sides by $2b$:
$$c = a \cdot \cos(\angle PEC)$$

7. **Using the Given Relationship $a^2 = b^2 + c^2$:**
- Since $a^2 = b^2 + c^2$, we know that $c = a \cdot \cos(\angle PEC)$.
- Therefore, $\cos(\angle PEC) = \frac{c}{a}$.

8. **Finding the Angle $\angle BPC$:**
- Since $\cos(\angle PEC) = \frac{c}{a}$ and $a^2 = b^2 + c^2$, we have:
$$\cos(\angle PEC) = \frac{c}{\sqrt{b^2 + c^2}}$$
- Given that $\cos(\angle PEC) = \frac{1}{\sqrt{2}}$, we conclude that $\angle PEC = 45^\circ$.

9. Conclusion:
- Since $\angle PEC = 45^\circ$ and $\angle BPE = 45^\circ$, the angle $\angle BPC = 90^\circ$.

$\blacksquare$