AMC 10 · 2015 · #10

Grade 8 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

polygon-inequalityperimeterinteger-pythagorean-triples systematic-enumerationbound-inequality-then-enumerate ↑ Prerequisites: polygon-inequalityperimeter

📏 Medium solution 💡 3 insights

Problem

Whole-number sides must be all different, the perimeter bounded, and no right angle allowed. Count the triangles.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Make a Systematic List

Every condition in the problem is a statement about three numbers, so the count of triangles turns into a count of integer triples. Listing them is the natural move, but a list only proves a count if nothing can hide outside it. So the real work is in two places the listing itself does not cover: bounding the search before listing, using the Extreme Principle, and then testing every survivor for a right angle rather than only recognizing the famous one.

1STEP 1

One triangle, one sorted triple

Sorting makes each triangle appear once.

noncongruent triangles ⟷ sorted integer triples (a,b,c), a ≤ b ≤ c

2STEP 2

Turn each word into an inequality

Every word becomes a plain inequality.

a ＜ b ＜ c, a+b ＞ c, a+b+c ≤ 14

3STEP 3

Prove the search has to stop

The search provably terminates.

a ≥ 4 → a+b+c ≥ 4+5+6 = 15 ＞ 14; a=1 → c ＜ b+1 → c ≤ b

4STEP 4

List both surviving cases in full

Only six triples survive the inequalities.

(2,3,4), (2,4,5), (2,5,6), (3,4,5), (3,4,6), (3,5,6)

5STEP 5

Test all six for a right angle

Exactly one of them has a right angle.

a²+b² vs c²: 13 ＜ 16, 20 ＜ 25, 29 ＜ 36, 25=25, 25 ＜ 36, 34 ＜ 36

6STEP 6

Count what is left

So 5 remain, choice (D).

6 - 1 = 5

Answer

Check the five survivors against every condition by hand. Perimeters are 9, 11, 13, 13, 14, all under 15. Every triple is strictly increasing, so none is isosceles or equilateral. The two shorter sides beat the longest in each: 2+3 ＞ 4, 2+4 ＞ 5, 2+5 ＞ 6, 3+4 ＞ 6, 3+5 ＞ 6, so all have positive area. And none satisfies a²+b²=c². The near-misses confirm the boundaries were read correctly: (2,3,5) fails because 2+3=5 collapses the triangle to zero area, (3,4,5) fails only because it is right, and (4,5,6) is scalene and not right yet has perimeter exactly 15, which "less than 15" excludes. Those last two are where the wrong answers live: keeping (4,5,6) gives 6, and keeping both (3,4,5) and (4,5,6) gives 7.

💡Key takeaway

A list only proves a count if you first prove the list has to end: here the shortest side cannot reach 4 without blowing the perimeter, so nothing can hide past a=3.

One triangle, one sorted triple
Turn each word into an inequality
Prove the search has to stop
List both surviving cases in full
Test all six for a right angle
Count what is left