2023 AMC 10B Problem 17

P16-20~3 minPrint

Algebraic Manipulation Solid Geometry Apply an identity: SFFT, sum of squares, difference of cubes, Vieta

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 $P$ ?

Show solution without answering

Key insight

The diagonal squared is a squared plus b squared plus c squared. That equals (a+b+c) squared minus 2(ab+bc+ca), and the edge total and face total give both parts.

Solution

We want the diagonal, and we are not asked for $a$ , $b$ , $c$ themselves. So look for a way to get the diagonal straight from the given totals.

1. The diagonal. Going across the bottom face gives a diagonal with square $a^2 + b^2$ (Pythagoras). Going up the height $c$ from there gives the long diagonal:
$\text{diagonal}^2 = a^2 + b^2 + c^2.$
Every interior diagonal of the box has this same length.
2. Edge total. A box has $4$ edges of each length, so $4(a + b + c) = 13$ and
$a + b + c = \frac{13}{4}.$
3. Face total. A box has $2$ faces of each shape, with areas $ab$ , $bc$ , $ca$ . So $2(ab + bc + ca) = \frac{11}{2}$ and
$ab + bc + ca = \frac{11}{4}.$
4. Connect them. Multiplying out a square of three terms gives
$(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca).$
(Check with $1, 2, 3$ : $6^2 = 36$ and $1 + 4 + 9 + 2 \cdot 11 = 36$ .)
5. Solve for what we want.
$a^2 + b^2 + c^2 = \left(\frac{13}{4}\right)^2 - 2 \cdot \frac{11}{4} = \frac{169}{16} - \frac{88}{16} = \frac{81}{16}.$
6. Take the square root. The diagonal is $\sqrt{\frac{81}{16}} = \frac{9}{4}$ .

We never needed the volume.

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

Why this works

The sum of squares $a^2 + b^2 + c^2$ can always be built from the sum $a + b + c$ and the sum of products $ab + bc + ca$ , using the square of a sum. When a problem gives totals like these and asks for something that treats $a$ , $b$ , $c$ the same way, try this identity before solving for each letter. Extra information, like the volume here, is sometimes not needed at all.

Alternative approach

You can also find the edges. The three numbers add to $\frac{13}{4}$ , their pairwise products add to $\frac{11}{4}$ , and their product (the volume) is $\frac{1}{2}$ . Only one set of three numbers has a given sum, sum of pairwise products and product, so if a guess fits all three facts, it is the answer. Try friendly values: $2$ , $1$ and $\frac{1}{4}$ .

Sum: $2 + 1 + \frac{1}{4} = \frac{13}{4}$ . Yes.
Pairwise products: $2 + \frac{1}{4} + \frac{1}{2} = \frac{11}{4}$ . Yes.
Product: $2 \cdot 1 \cdot \frac{1}{4} = \frac{1}{2}$ . Yes.

So the edges are $2$ , $1$ , $\frac{1}{4}$ , and the diagonal is $\sqrt{4 + 1 + \frac{1}{16}} = \sqrt{\frac{81}{16}} = \frac{9}{4}$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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

Review in the AoPS books

Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Algebra, Chapter 22: Special Manipulations
Introduction to Geometry, Chapter 15: Curved Surfaces