2000 AMC 10 Problem 10

P1-10~3 minPrint

Triangles: Area & Pythagorean Absolute Value & Inequalities Bound the quantity above/below or estimate to pin it down Consider the largest/smallest element or boundary case

The sides of a triangle with positive area have lengths $4$ , $6$ , and $x$ . The sides of a second triangle with positive area have lengths $4$ , $6$ , and $y$ . What is the smallest positive number that is not a possible value of $|x-y|$ ?

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Key insight

The third side must be strictly between 2 and 10. Two numbers in that range can differ by anything less than 8, but never by exactly 8.

Solution

1. Which lengths can $x$ be? In a real triangle, any two sides together must be longer than the third side.
- $4 + 6 > x$ , so $x < 10$ .
- $4 + x > 6$ , so $x > 2$ .

So $x$ is strictly between $2$ and $10$ . The same is true for $y$ .
2. How far apart can $x$ and $y$ be? The farthest apart they can be is when one is near $10$ and the other is near $2$ . But neither can actually reach $10$ or $2$ , so the gap is always less than $10 - 2 = 8$ . The gap $8$ is impossible.
3. Every smaller gap is possible. For example:
- gap $2$ : $x = 5$ , $y = 3$ .
- gap $4$ : $x = 7$ , $y = 3$ .
- gap $6$ : $x = 9$ , $y = 3$ .
- gap $7.9$ : $x = 9.95$ , $y = 2.05$ .

Any gap below $8$ works the same way: put $y$ a little above $2$ and $x$ that much higher.
4. So $2$ , $4$ and $6$ are all possible, and $8$ is the smallest positive number that is not.

The answer is $\boxed{\textbf{(D)}\ 8}$ .

Why this works

The rule "any two sides add to more than the third" gives a range for the unknown side. The words "positive area" mean the triangle cannot be flat, so the ends of the range are not allowed. Two numbers from a range with the ends left out can get close to the full width of the range, but never reach it. Watching "less than" versus "less than or equal to" is the whole problem.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

For a triangle with positive area and sides $4$ , $6$ , $x$ , the triangle inequality must hold strictly: $x < 4 + 6 = 10$ and $x + 4 > 6$ , so $x > 2$ . Thus $2 < x < 10$ , and likewise $2 < y < 10$ .

Two numbers from the open interval $(2, 10)$ differ by less than $10 - 2 = 8$ , so $|x - y| < 8$ ; the value $8$ is impossible. On the other hand every value in $[0, 8)$ occurs: for a target $d < 8$ , take $y$ slightly above $2$ and $x = y + d$ , which is still below $10$ . In particular $2$ , $4$ , $6$ are all attainable (for instance $x = 9, y = 3$ gives $6$ ).

Therefore the smallest positive number that $|x - y|$ cannot equal is $8$ .

The answer is $\boxed{\textbf{(D)}\ 8}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2)
Introduction to Algebra, Chapter 20: Special Functions (§20.2, §20.5)
Introduction to Geometry, Chapter 6: Right Triangles (§6.1, §6.2, §6.3)
Introduction to Algebra, Chapter 15: More Inequalities (§15.1, §15.3, §15.4)