Four sides and a diagonal of a quadrilateral are of lengths $10,20,28,50,75$, not necessarily in that order. Which amongst them is the only possible length of the diagonal?

Step 1: Let the diagonal be $d$. The diagonal splits the quadrilateral into two triangles with sides $(x,y,d)$ and $(u,v,d)$. By the triangle inequality:

Also, the sum of all four sides must be strictly greater than $2d$ (since the perimeter is greater than twice the diagonal path).

Step 2: Let's test the possible choices for $d$ from the set $\{10,20,28,50,75\}$:

• Case 1: $d=75$. The four remaining elements are $10,20,28,50$. Their sum is $10+20+28+50=108$. But we must have $x+y>75$ and $u+v>75$, meaning the sum of the four sides must be at least $2d=150$. Since $108<150$, $75$ cannot be the diagonal.

• Case 2: $d=50$. The remaining elements are $10,20,28,75$. One of the triangles must contain the side of length $75$, say $(75,x,50)$. By the triangle inequality, $75<50+x⟹x>25$. The only available value in $\{10,20,28\}$ greater than $25$ is $28$. Thus, one triangle has sides $(75,28,50)$, which is valid ($75-28=47<50<78$). The other triangle must have sides $(10,20,50)$, but $10+20=30<50$, violating the triangle inequality. Thus, $50$ is invalid.

• Case 3: $d=20$. The remaining elements are $10,28,50,75$. To pair them, the side $75$ must be in a triangle with $50$ and $20$ (since no other side is large enough to satisfy $75<x+20$). But $50+20=70<75$, violating the inequality. So $20$ is invalid.

• Case 4: $d=10$. Similarly, the side $75$ would need to be paired with $50$ and $10$, but $50+10=60<75$, violating the inequality. So $10$ is invalid.

• Case 5: $d=28$. The remaining elements are $10,20,50,75$. We can form the triangles with sides $(10,20,28)$ (valid: $10+20=30>28$) and $(50,75,28)$ (valid: $50+28=78>75$). Both triangle inequalities are satisfied.

Thus, the only possible length of the diagonal is $28$.