AMC 10 · 2019 · #15
Grade 8 geometry-2dProblem
Right triangles T1 and T2, have areas of 1 and 2, respectively. A side of T1 is congruent to a side of T2, and a different side of T1 is congruent to a different side of T2. What is the square of the product of the lengths of the other (third) sides of T1 and T2?
Pick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two right triangles $T_1, T_2$ have areas $1$ and $2$. One side of $T_1$ equals one side of $T_2$, and a different side of $T_1$ equals a different side of $T_2$. Find the square of the product of the two not-shared (third) sides.
Givens: $T_1$ is a right triangle with area $1$; $T_2$ is a right triangle with area $2$; Two pairs of sides match: one side of $T_1$ $=$ one side of $T_2$; another side of $T_1$ $=$ another side of $T_2$; The third side of $T_1$ and the third side of $T_2$ may differ — and they do; Answer choices: $\tfrac{28}{3},\, 10,\, \tfrac{32}{3},\, \tfrac{34}{3},\, 12$
Unknowns: $(\text{third side of } T_1)^2 \cdot (\text{third side of } T_2)^2$
Understand
Restated: Two right triangles $T_1, T_2$ have areas $1$ and $2$. One side of $T_1$ equals one side of $T_2$, and a different side of $T_1$ equals a different side of $T_2$. Find the square of the product of the two not-shared (third) sides.
Givens: $T_1$ is a right triangle with area $1$; $T_2$ is a right triangle with area $2$; Two pairs of sides match: one side of $T_1$ $=$ one side of $T_2$; another side of $T_1$ $=$ another side of $T_2$; The third side of $T_1$ and the third side of $T_2$ may differ — and they do; Answer choices: $\tfrac{28}{3},\, 10,\, \tfrac{32}{3},\, \tfrac{34}{3},\, 12$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #9 Solve an Easier Related Problem, #13 Convert to Algebra, #3 Eliminate Possibilities
Tool #1 (Draw): sketch two right triangles with labeled legs/hypotenuse and discover the only role-swap consistent with both areas. Tool #9 (Easier Problem): once we recognize the shared values are a leg of $T_1$ paired with the same as a leg of $T_2$, and the hypotenuse of $T_1$ paired with a leg of $T_2$ — the rest is two area equations in two unknowns. Tool #13 (Algebra): set up $ab = 4$ and $a^2(b^2-a^2) = 4$, then evaluate $b^4 - a^4$ — no heavy machinery needed. Tool #3 matches $\tfrac{28}{3}$ to choice $(A)$.
Execute — Answer: A
8.G.B.7 Step 1 - Draw $T_1$ and $T_2$.
- Call the two shared lengths $a$ and $b$ with $a < b$.
- In a right triangle the hypotenuse is the longest side.
- Try: $a, b$ are both legs of $T_2$ (so hypotenuse of $T_2$ is something new); but in $T_1$, $a$ is a leg and $b$ is the hypotenuse (so the other leg of $T_1$ is something new).
- This is the only role-assignment that lets the two triangles be non-congruent while sharing $a$ and $b$.
💡 Grade 8 Pythagoras: the labeled diagram pins down which length plays hypotenuse where.
6.G.A.1 Step 2 - Write the two area equations.
- $T_1$ has legs $a$ and $\sqrt{b^2-a^2}$, area $\tfrac{1}{2}a\sqrt{b^2-a^2} = 1$.
- $T_2$ has legs $a, b$, area $\tfrac{1}{2} a b = 2$.
💡 Grade 6 area: a right triangle's area is half the product of its two legs.
8.EE.A.2 Step 3 - Simplify: from $T_2$'s area $ab = 4$.
- From $T_1$'s area, square both sides: $a^2(b^2 - a^2) = 4$.
💡 Grade 8 squares: squaring removes the radical from the $T_1$ area equation.
8.EE.A.1 Step 4 Expand the second equation and substitute $a^2 b^2 = (ab)^2 = 16$ from the first.
💡 Grade 8 exponents: $a^2 b^2 = (ab)^2$ collapses the unknown into one fourth-power equation.
8.EE.A.1 Step 5 Find $b^4$ from $b = 4/a$, so $b^4 = 256/a^4 = 256/12 = 64/3$.
💡 Grade 8: once $ab$ and $a^4$ are known, $b^4$ falls out by a single division.
8.EE.A.2 Step 6 - The non-shared sides are $\sqrt{b^2 - a^2}$ (third side of $T_1$) and $\sqrt{a^2 + b^2}$ (third side of $T_2$).
- Their product squared equals the product of the two radicands.
💡 Grade 8: a difference of squares neatly turns the product into $b^4 - a^4$.
5.NF.A.1 Step 7 Substitute $b^4 = 64/3$ and $a^4 = 12$.
💡 Grade 5 fractions: subtract with a common denominator.
5.NF.A.1 Step 8 Match $\tfrac{28}{3}$ to the choices.
💡 Grade 5: the answer matches the first fractional choice.
8.G.B.7 Draw $T_1$ and $T_2$. Call the two shared lengths $a$ and $b$ with $a < b$. In a 6.G.A.1 Write the two area equations. $T_1$ has legs $a$ and $\sqrt{b^2-a^2}$, area $\tf 8.EE.A.2 Simplify: from $T_2$'s area $ab = 4$. From $T_1$'s area, square both sides: $a^2 8.EE.A.1 Expand the second equation and substitute $a^2 b^2 = (ab)^2 = 16$ from the first 8.EE.A.1 Find $b^4$ from $b = 4/a$, so $b^4 = 256/a^4 = 256/12 = 64/3$. 8.EE.A.2 The non-shared sides are $\sqrt{b^2 - a^2}$ (third side of $T_1$) and $\sqrt{a^2 5.NF.A.1 Substitute $b^4 = 64/3$ and $a^4 = 12$. 5.NF.A.1 Match $\tfrac{28}{3}$ to the choices. Review
Reasonableness: Solve numerically: $a^4 = 12 \Rightarrow a^2 = 2\sqrt{3}$, $b^2 = 16/a^2 = 16/(2\sqrt{3}) = 8/\sqrt{3}$. Other leg of $T_1$ $= \sqrt{b^2 - a^2} = \sqrt{8/\sqrt{3} - 2\sqrt{3}} = \sqrt{(8 - 2 \cdot 3)/\sqrt{3}} = \sqrt{2/\sqrt{3}}$. Hypotenuse of $T_2$ $= \sqrt{a^2 + b^2} = \sqrt{2\sqrt{3} + 8/\sqrt{3}} = \sqrt{(6 + 8)/\sqrt{3}} = \sqrt{14/\sqrt{3}}$. Their product squared $= (2/\sqrt{3})(14/\sqrt{3}) = 28/3$ ✓. Sanity numerics: $a \approx 1.86, b \approx 2.15$, so $T_2$ area $\approx \tfrac{1}{2}(1.86)(2.15) \approx 2.00$ ✓ and $T_1$ area $\approx \tfrac{1}{2}(1.86)\sqrt{2.15^2 - 1.86^2} \approx \tfrac{1}{2}(1.86)(1.07) \approx 1.00$ ✓.
Alternative: Tool #2 (Systematic list of role-assignments): enumerate the ways two right triangles can share two sides. (i) Both legs of each — then they are congruent (same hypotenuse via Pythagoras), areas equal — fails. (ii) Leg + hypotenuse of each — then by Pythagoras both have the same third side too — congruent again. (iii) The asymmetric case used above: in one triangle the shared values are a leg + hypotenuse; in the other they are two legs. Only case (iii) admits area $1$ vs $2$ — same setup, same answer $(A)$.
CCSS standards used (min grade 8)
5.NF.A.1Add and subtract fractions with unlike denominators (Computing $\tfrac{64}{3} - 12 = \tfrac{28}{3}$ in the final step.)6.G.A.1Find area of triangles, special quadrilaterals, and polygons by composing (Writing each right-triangle area as half the product of legs.)8.EE.A.1Know and apply the properties of integer exponents (Using $(ab)^2 = a^2 b^2$ and computing $b^4 = (ab)^4 / a^4$.)8.EE.A.2Use square root and cube root symbols to represent solutions (Squaring the $T_1$ area equation and applying difference of squares $b^4 - a^4 = (b^2-a^2)(b^2+a^2)$.)8.G.B.7Apply the Pythagorean theorem to determine unknown side lengths in right triangles (Identifying the hypotenuse-vs-leg roles in $T_1, T_2$ and the resulting third sides $\sqrt{b^2 - a^2}$ and $\sqrt{a^2 + b^2}$.)
⭐ This AMC 10 problem only needs Grade 8 Pythagoras you already know! The shared sides $a, b$ have to play different roles in the two triangles: legs of $T_2$, but leg-and-hypotenuse of $T_1$. The two area equations give $ab = 4$ and $a^2(b^2 - a^2) = 4$, which yield $a^4 = 12$ and $b^4 = 64/3$. The squared product of the third sides is $(b^2-a^2)(b^2+a^2) = b^4 - a^4 = \mathbf{\tfrac{28}{3}}$, answer $(A)$.
⭐ This AMC 10 problem only needs Grade 8 Pythagoras you already know! The shared sides $a, b$ have to play different roles in the two triangles: legs of $T_2$, but leg-and-hypotenuse of $T_1$. The two area equations give $ab = 4$ and $a^2(b^2 - a^2) = 4$, which yield $a^4 = 12$ and $b^4 = 64/3$. The squared product of the third sides is $(b^2-a^2)(b^2+a^2) = b^4 - a^4 = \mathbf{\tfrac{28}{3}}$, answer $(A)$.
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