Problem 1479

### Part (a)

1. Define the segments:
Let $AN = y$, $MN = x$, $PM = t$, and $AP = z$.

2. Use the Pythagorean theorem:
Since $\Delta ANM$ and $\Delta APM$ are right triangles sharing a hypotenuse $AM$, we have:
$$x^2 + y^2 = z^2 + t^2$$
This can be rewritten as:
$$(x - z)(x + z) = (t - y)(t + y)$$

3. Apply Pitot's theorem:
Since $MNAP$ is circumscribed, by Pitot's theorem, the sum of the lengths of opposite sides is equal:
$$x + z = y + t$$

4. Combine the equations:
Using the two equalities:
$$(x - z)(x + z) = (t - y)(t + y)$$
and
$$x + z = y + t$$
we can substitute $x + z$ with $y + t$ in the first equation:
$$(x - z)(y + t) = (t - y)(y + t)$$

5. Simplify the equation:
Since $y + t \neq 0$, we can divide both sides by $y + t$:
$$x - z = t - y$$

6. **Solve for $x$ and $t$:**
Adding the two equations:
$$x + z = y + t$$
and
$$x - z = t - y$$
we get:
$$2x = 2t \implies x = t$$
Similarly, subtracting the two equations:
$$x + z = y + t$$
and
$$x - z = t - y$$
we get:
$$2z = 2y \implies z = y$$

7. Conclusion:
Since $x = t$ and $y = z$, $MNAP$ is a kite.

$\blacksquare$

### Part (b)

1. Assume the trapezoid:
Suppose there is such a trapezoid and label it $ABCD$ (with $AB \parallel CD$ and $CD > AB$). Let $AD = BC = x$, $AB = y$, and $CD = z$.

2. Apply Pitot's theorem:
Since $ABCD$ is circumscribed, by Pitot's theorem, we must have:
$$y + z = 2x$$
Squaring both sides:
$$(y + z)^2 = 4x^2$$

3. Use the orthodiagonal property:
Since $ABCD$ is orthodiagonal, a well-known lemma gives that:
$$y^2 + z^2 = 2x^2$$
Doubling both sides:
$$2(y^2 + z^2) = 4x^2$$

4. Combine the equations:
Equating the two expressions for $4x^2$:
$$(y + z)^2 = 2(y^2 + z^2)$$
Expanding and simplifying:
$$y^2 + 2yz + z^2 = 2y^2 + 2z^2 \implies y^2 + z^2 + 2yz = 2y^2 + 2z^2 \implies 2yz = y^2 + z^2$$
Rearranging:
$$(y - z)^2 = 0 \implies y = z$$

5. Conclusion:
This implies that $AB = CD$, which contradicts the assumption that $ABCD$ is a trapezoid (since $AB \parallel CD$ and $CD > AB$). Therefore, it is not possible for an isosceles trapezoid to be both orthodiagonal and circumscribed.

$\blacksquare$