AMC 10 · 2014 · #13

Grade 9 geometry-2d

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

polygon-inequalityquadratic-equationsoptimizationfraction-arithmetic extreme-principleconvert-to-algebra ↑ Prerequisites: polygon-inequalityquadratic-equations

📏 Medium solution 💡 3 insights

Problem

Neither of two triples of lengths may form a real triangle. Find the smallest possible larger number.

Pick an answer.

(A)

$\frac{3+\sqrt{3}}{2}$

(B)

$\frac{5}{2}$

(C)

$\frac{3+\sqrt{5}}{2}$

(D)

$\frac{3+\sqrt{6}}{2}$

(E)

How to solve

Strategy Extreme Principle

The problem is stated as a ban: two triples are forbidden from being triangles. A ban is hard to compute with, so the first move is to flip it. Three lengths make a triangle with positive area exactly when the longest is shorter than the sum of the other two, so "no triangle with positive area" is exactly "the longest is at least the sum of the other two". That turns each ban into a plain inequality. The two inequalities involve both a and b, and only b is being minimised, so I reorganise them: I hold b fixed and ask which values of a survive. One condition pushes a down, the other pushes a up, so together they trap a inside an interval. A value of b is legal exactly when that interval is not empty, which is a single inequality in b alone. Its boundary is where the interval shrinks to one point, and that is the extreme case the question is asking about. The last piece is the part that is easy to skip: proving the boundary value is actually attained, by producing the exact a that goes with it and checking both original conditions on the nose.

1STEP 1

Flip the ban into an inequality

Failing is the triangle inequality reversed.

positive area ⇔ longest ＜ sum of other two ⟹ no positive area ⇔ longest ≥ sum of other two

2STEP 2

Write both bans as inequalities

Both bans become clean inequalities.

b ≥ 1 + a, 1/a + 1/b ≤ 1, b ＞ 2

3STEP 3

Clear fractions in the second ban

Clearing fractions makes the second one usable.

1/a + 1/b ≤ 1 ⟺ a + b ≤ ab ⟺ b ≤ a(b-1) ⟺ a ≥ b/(b-1)

4STEP 4

Trap a between two bounds

The smaller number is trapped between two bounds.

b/(b-1) ≤ a ≤ b-1, b/(b-1) ＞ 1 and b-1 ＜ b hold automatically

5STEP 5

Non-empty interval forces a quadratic

A non-empty trap forces one quadratic.

b/(b-1) ≤ b-1 ⇔ b² - 3b + 1 ≥ 0 ⇔ b ≤ (3-√(5))/2 or b ≥ (3+√(5))/2; b ＞ 2 → b ≥ (3+√(5))/2

6STEP 6

Show the bound is reached

Its root is reached, so choice (B) stands.

a = (1+√(5))/2, b = (3+√(5))/2: 1 + a = b, 1/a + 1/a² = (a+1)/a² = 1

Answer

(3+√(5))/2

Test the runner-up choices directly. For b = 5/2 the trap from step 4 asks for a with 2.5/1.5 = 1.6 ≤ a ≤ 1.5, which is empty, so choice (B) is impossible; the same check kills (3+√(3))/2 ≈ 2.37, whose interval is [1.73, 1.37]. For b = 3 the interval is [1.5, 2], which works, so 3 is legal but not smallest, and every choice above (3+√(5))/2 is legal for the same reason. That ordering is a good sign: legality is a tail condition, true for all large b and false below a single threshold, so the answer must be exactly that threshold, ≈ 2.618, sitting between the failing 2.5 and the working 2.724. It is also worth naming what the phrase "positive area" is doing. If the problem had banned even the flat case, the conditions would be strict, b ＞ 1 + a and 1/a + 1/b ＜ 1, the legal set would be b ＞ (3+√(5))/2 with no smallest element, and the question would have no answer. The wording is what makes step 6 possible.

💡Key takeaway

Rewrite "these lengths cannot make a triangle" as "the longest side is at least the other two combined", then let one rule push the middle number up and the other push it down; the smallest b is where the two pushes exactly meet.

Flip the ban into an inequality
Write both bans as inequalities
Clear fractions in the second ban
Trap a between two bounds
Non-empty interval forces a quadratic
Show the bound is reached