Problem 1715

### Part (a)
1. Let $h$ be the altitude corresponding to the side $MN$ in triangle $MNP$.
2. By the properties of triangles, the area $A$ of $\triangle MNP$ can be expressed as:
$$A = \frac{1}{2} \times MN \times h$$
3. Using Heron's formula, the area $A$ can also be expressed in terms of the sides $MN, MP,$ and $NP$. Let $a = MN$, $b = MP$, and $c = NP$. The semi-perimeter $s$ is:
$$s = \frac{a + b + c}{2}$$
Then, the area $A$ is:
$$A = \sqrt{s(s-a)(s-b)(s-c)}$$
4. Equating the two expressions for the area, we get:
$$\frac{1}{2} \times a \times h = \sqrt{s(s-a)(s-b)(s-c)}$$
Solving for $h$:
$$h = \frac{2 \sqrt{s(s-a)(s-b)(s-c)}}{a}$$
5. Given that $a, b, c < 2$, we need to show that:
$$h < \sqrt{4 - \frac{a^2}{4}}$$
6. Since $a, b, c < 2$, the maximum possible value for $s$ is:
$$s < 3$$
7. To find an upper bound for $h$, we use the fact that the maximum area of a triangle with sides less than 2 is achieved when the triangle is equilateral with side length approaching 2. For an equilateral triangle with side length $a$:
$$A = \frac{\sqrt{3}}{4} a^2$$
The altitude $h$ in this case is:
$$h = \frac{\sqrt{3}}{2} a$$
8. For $a < 2$, we have:
$$h < \frac{\sqrt{3}}{2} \times 2 = \sqrt{3}$$
9. We need to show that:
$$\sqrt{3} < \sqrt{4 - \frac{a^2}{4}}$$
Squaring both sides:
$$3 < 4 - \frac{a^2}{4}$$
Simplifying:
$$\frac{a^2}{4} < 1$$
$$a^2 < 4$$
Which is true since $a < 2$.

Thus, the altitude $h$ corresponding to the side $MN$ is less than $\sqrt{4 - \frac{MN^2}{4}}$.

### Part (b)
1. Let $ABCD$ be a tetrahedron with at least 5 edges having lengths less than 2.
2. Without loss of generality, assume $CD = a < 2$ and $AB$ is not necessarily less than 2.
3. Denote $M$ as the projection of point $B$ on the line $CD$. In $\triangle BCD$, the sides are less than 2, so:
$$BM < \sqrt{4 - \frac{a^2}{4}}$$
4. Similarly, the altitude $h$ from vertex $A$ in $\triangle ACD$ is less than:
$$\sqrt{4 - \frac{a^2}{4}}$$
5. Therefore, the volume $V$ of tetrahedron $ABCD$ is:
$$V = \frac{1}{3} \times h \times \text{Area of } \triangle BCD$$
6. The area of $\triangle BCD$ is maximized when $BCD$ is an equilateral triangle with side length approaching 2. The maximum area is:
$$\text{Area of } \triangle BCD = \frac{\sqrt{3}}{4} a^2$$
7. Thus, the volume $V$ is:
$$V < \frac{1}{3} \times \sqrt{4 - \frac{a^2}{4}} \times \frac{\sqrt{3}}{4} a^2$$
8. Simplifying, we need to show:
$$V < 1$$
9. Given $a < 2$, we have:
$$\frac{a(16 - a^2)}{24} < 1$$
10. To prove this, consider the polynomial:
$$a^3 - 16a + 24 = (a - 2)(a^2 + 2a - 12)$$
Since $a < 2$ and $a^2 + 2a < 12$, the inequality holds.

Thus, the volume of the tetrahedron is less than 1.