AMC 10 · 2021 · #2

Grade 9 algebra
Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination
signed-square-rootzero-product-propertysign-analysisabsolute-value easier-related-problemextreme-principle ↑ Prerequisites: signed-square-root
📏 Medium solution 💡 2 insights
Problem
For two real numbers, decide exactly when the square root of the sum of their squares equals their sum. Each answer choice states a candidate condition; the right one must hold for every pair making the equation true, and must force the equation in return.

Pick an answer.

(A)
It is never true.
(B)
$\text{It is true if and only if } ab=0.$
(C)
$\text{It is true if and only if } a+b\ge 0.$
(D)
$\text{It is true if and only if } ab=0 \text{ and } a+b\ge 0.$
(E)
It is always true.
How to solve
Strategy Solve an Easier Related Problem

A radical equation is hard to compare term by term, so Tool #9 (Solve an Easier Related Problem) replaces √(a²+b²)=a+b with the squared equation a²+b²=(a+b)², which is a polynomial statement anyone can expand. Squaring is a one-way street, though, so Tool #14 (Extreme Principle) supplies the missing half: the smallest value a principal square root can take is 0, so the right side is forced to be nonnegative, and that is exactly the condition that makes squaring reversible. With both halves in hand the true condition is known, and Tool #3 (Eliminate Possibilities) confirms it by producing one concrete counterexample against each of the other four choices.

1STEP 1

The left side is never negative

The left side is never negative.

√(a²+b²) ≥ 0 ⟹ a+b ≥ 0
2STEP 2

Square to remove the radical

Squaring leaves the condition that the product is zero.

a²+b²=(a+b)²=a²+2ab+b² ⟹ 0=2ab ⟹ ab=0
3STEP 3

Check the condition is also enough

Check the condition is also sufficient.

b=0, a ≥ 0 ⟹ √(a²+0²)=√(a²)=|a|=a=a+b
4STEP 4

Kill the other four choices

Counterexamples leave product zero and sum nonnegative.

(3,0)→(A) out; (-1,0)→(B) out; (1,1)→(C),(E) out ⟹ (D)
Answer
It is true if and only if ab=0 and a+b ≥ 0.
The condition in (D) describes the set of points where the equation holds as the nonnegative part of the a-axis together with the nonnegative part of the b-axis: ab=0 puts the point on an axis, and a+b ≥ 0 keeps only the half of each axis that points away from the origin in the positive direction. Spot-checking that picture works: (5,0) gives √(25)=5, true; (0,7) gives √(49)=7, true; (0,0) gives 0=0, true; (0,-2) is on an axis but has a+b=-2 ＜ 0 and indeed √(4)=2 ≠ -2; and (3,4) is off the axes with √(25)=5 ≠ 7. Every case matches (D) and no other choice describes that same set.
💡Key takeaway

Squaring both sides of a radical equation only tells half the story — you also have to demand that the other side is not negative, because a square root never is.

  • The left side is never negative
  • Square to remove the radical
  • Check the condition is also enough
  • Kill the other four choices

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