Competition · AMC preparation · step 4 of 4
AMC 10 · 2019B · #15
Grade 8 geometry-2dPick an answer.
AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 T₁ paired with the same as a leg of T₂, and the hypotenuse of T₁ paired with a leg of T₂ — the rest is two area equations in two unknowns. Tool #13 (Algebra): set up ab = 4 and a²(b²-a²) = 4, then evaluate b⁴ - a⁴ — no heavy machinery needed. Tool #3 matches 28/3 to choice (A).
Draw both triangles
Draw both triangles, shared sides a < b. Only one role-swap works: a and b are legs of T₂; in T₁, a is a leg and b the hypotenuse.
Grade 8 Pythagoras: the labeled diagram pins down which length plays hypotenuse where.
8.G.B.7Draw A DiagramWrite the two area equations
Write the area equations: T₁ has legs a and √(b² - a²) so ½a√(b² - a²) = 1; T₂ has legs a, b so ½ab = 2.
Grade 6 area: a right triangle's area is half the product of its two legs.
6.G.A.1Convert To AlgebraSimplify the second area
Simplify: T₂'s area gives ab = 4; squaring T₁'s area gives a²(b² - a²) = 4.
Grade 8 squares: squaring removes the radical from the T₁ area equation.
8.EE.A.2Convert To AlgebraSubstitute to find a fourth power
Expand a²b² - a⁴ = 4 and substitute a²b² = (ab)² = 16, so a⁴ = 12.
Grade 8 exponents: a² b² = (ab)² collapses the unknown into one fourth-power equation.
Two squares multiplied together are the square of the product, which collapses the unknowns into one.
▸ Why?
An exponent counts how many times a factor is used, so squares combine by using each factor twice.
▸ Why?
The factors can be reordered freely, which is what lets the two squares regroup into one.
Find the other fourth power
From b = , b⁴ = = = .
Grade 8: once ab and a⁴ are known, b⁴ falls out by a single division.
8.EE.A.1Convert To AlgebraMultiply the two unshared sides
The third sides are √(b² - a²) and √(a² + b²); their product squared is (b² - a²)(b² + a²) = b⁴ - a⁴.
Grade 8: a difference of squares neatly turns the product into b⁴ - a⁴.
8.EE.A.2Convert To AlgebraSubstitute the fourth powers
Substitute b⁴ = and a⁴ = 12: b⁴ - a⁴ = - 12 = .
Grade 5 fractions: subtract with a common denominator.
5.NF.A.1Convert To AlgebraMatch against the choices
Match to choice (A).
Grade 5: the answer matches the first fractional choice.
5.NF.A.1Eliminate PossibilitiesThis 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 T₂, but leg-and-hypotenuse of T₁. The two area equations give ab = 4 and a²(b² - a²) = 4, which yield a⁴ = 12 and b⁴ = . The squared product of the third sides is (b²-a²)(b²+a²) = b⁴ - a⁴ = , answer (A).
- Draw both triangles
- Write the two area equations
- Simplify the second area
- Substitute to find a fourth power
- Find the other fourth power
- Multiply the two unshared sides
- Substitute the fourth powers
- Match against the choices
A parent dashboard for the family lives at sensimlab.com.