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2023 AMC 10B Problem 17

2023 AMC 10B Problem 17AMC 10 Step by Step

Name: Date: September 13, 2026

Algebraic Manipulation · difficulty 3 of 5 · about 3 min · position P16-20

A rectangular box $P$ has distinct edge lengths $a$ , $b$ , and $c$ . The sum of the lengths of all $12$ edges of $P$ is $13$ , the sum of the areas of all $6$ faces of $P$ is $\dfrac{11}{2}$ , and the volume of $P$ is $\dfrac{1}{2}$ . What is the length of the longest interior diagonal connecting two vertices of $P$ ?

A)
$2$
B)
$\frac{3}{8}$
C)
$\frac{9}{8}$
D)
$\frac{9}{4}$
E)
$\frac{3}{2}$

Answer

Key insight

The diagonal is sqrt(a^2+b^2+c^2) = sqrt((a+b+c)^2 - 2(ab+bc+ca)); the edge sum and face sum give both pieces without finding a, b, c.

Solution

A box has four edges of each length, so $4(a + b + c) = 13$ and $a + b + c = \frac{13}{4}$ . It has two faces of each of the three shapes, so $2(ab + bc + ca) = \frac{11}{2}$ and $ab + bc + ca = \frac{11}{4}$ .

Every interior diagonal has length $\sqrt{a^2 + b^2 + c^2}$ , and the identity $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ gives
$a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - 2 \cdot \frac{11}{4} = \frac{169}{16} - \frac{88}{16} = \frac{81}{16}.$
So the diagonal is $\sqrt{81/16} = \frac{9}{4}$ . The volume is never needed.

The answer is $\boxed{\textbf{(D)}\ \frac{9}{4}}$ .

Why this works

Edge sum, face-area sum and volume are the three elementary symmetric functions of $a, b, c$ , and any symmetric quantity, such as $a^2 + b^2 + c^2$ , can be written in terms of them. Recognizing that the question asks for a symmetric expression saves you from solving a cubic. Extra data (here the volume and "distinct") is often a decoy.

Alternative approach

With all three symmetric functions, $a, b, c$ are the roots of $x^3 - \frac{13}{4}x^2 + \frac{11}{4}x - \frac{1}{2} = 0$ , i.e. $4x^3 - 13x^2 + 11x - 2 = (x - 2)(4x - 1)(x - 1) = 0$ . The edges are $2, 1, \frac{1}{4}$ and the diagonal is $\sqrt{4 + 1 + \frac{1}{16}} = \frac{9}{4}$ .

The trap

Using 13 as a + b + c (each edge length appears 4 times, so a + b + c = 13/4) or 11/2 as ab + bc + ca instead of 11/4.

Common mistakes

Using 13 as a + b + c (each edge length appears 4 times, so a + b + c = 13/4) or 11/2 as ab + bc + ca instead of 11/4.
Trying to solve for the individual edge lengths first; it works, but the cubic costs time and invites arithmetic slips.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta