AMC 10 · 2023 · #17
Grade 8 geometry-3dPick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
We never need the individual edges — only the symmetric sums s₁=a+b+c and s₂=ab+bc+ca. Tool #7 (Identify Subproblems) chops the job into three rungs: (a) read off s₁ and s₂ from the given totals, (b) plug into the identity (s₁)² = a²+b²+c² + 2s₂ to get a²+b²+c², (c) take the square root via the Pythagorean-theorem extension to find the space diagonal. Tool #13 (Convert to Algebra) handles the identity; Tool #8 (Analyze the Units) confirms the answer is a length (not an area or volume).
Read the two symmetric sums off the totals: 4(a+b+c)=13 gives a+b+c=, and the six faces give 2(ab+bc+ca)=.
A rectangular box's surface area and edge total are themselves "sum of edges" and "sum of face products" — exactly the two symmetric sums of a,b,c we need.
6.G.A.4Identify SubproblemsThe identity (a+b+c)² = a²+b²+c² + 2(ab+bc+ca) gives a²+b²+c² = ()² - = ; the volume is a red herring.
Squaring a+b+c already contains every a²,b²,c² once plus every cross term ab,bc,ca twice — exactly what we need to subtract off.
8.EE.A.2Convert To AlgebraThe space diagonal is d=√(a²+b²+c²) by the 3D Pythagorean theorem; substituting gives d=√()=.
The space diagonal of a box is the hypotenuse of a right triangle whose legs are a face diagonal and the third edge — applying Pythagoras twice gives d² = a² + b² + c².
8.G.B.7Identify SubproblemsUnits check: (a+b+c)² and 2(ab+bc+ca) are both areas, so their difference is an area, and its square root is a length — a diagonal.
Tracking units catches errors fast: a diagonal must come out as a length, not a number times an area.
5.MD.A.1Analyze The UnitsYou never need the individual edges a,b,c: the totals give a+b+c= and 2(ab+bc+ca)=, and the identity (a+b+c)² = a²+b²+c² + 2(ab+bc+ca) delivers a²+b²+c² = . The space diagonal is √() = (D) — and the volume was a red herring.