Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2010 USAJMO Problems/Problem 4

Revision as of 20:54, 14 April 2014 by Suli (talk | contribs) (→‎Solution 2)

Problem

A triangle is called a parabolic triangle if its vertices lie on a parabola $y = x^2$. Prove that for every nonnegative integer $n$, there is an odd number $m$ and a parabolic triangle with vertices at three distinct points with integer coordinates with area $(2^nm)^2$.

A Small Hint

Before you read the solution, try using induction on n. (And don't step by one!)

Solution

Let the vertices of the triangle be $(a, a^2), (b, b^2), (c, c^2)$. The area of the triangle is the absolute value of $A$ in the equation:

\[A = \frac{1}{2}\det\left\vert \begin{array}{c c} b-a & c - a\\ b^2 - a^2 & c^2 - a^2 \end{array}\right\vert = \frac{(b-a)(c-a)(c-b)}{2}\]

If we choose $a < b < c$, $A > 0$ and gives the actual area. Furthermore, we clearly see that the area does not change when we subtract the same constant value from each of $a$, $b$ and $c$. Thus, all possible areas can be obtained with $a = 0$, in which case $A = \frac{1}{2}bc(c-b)$.

If a particular choice of $b$ and $c$ gives an area $A = (2^nm)^2$, with $n$ a positive integer and $m$ a positive odd integer, then setting $b' = 4b$, $c' = 4c$ gives an area $A' = 4^3 A = 8^2 A = (2^{n+3}m)^2$.

Therefore, if we can find solutions for $n = 0$, $n = 1$ and $n = 2$, all other solutions can be generated by repeated multiplication of $b$ and $c$ by a factor of $4$.

Setting $b=1$ and $c=2$, we get $A=1 = (2^0\cdot1)^2$, which yields the $n=0$ case.

Setting $b=1$ and $c=9$, we get $A = 9\cdot4 = (2^1\cdot3)^2$, which yields the $n=1$ case.

Setting $b=1$ and $c=5$, we get $A=1\cdot2\cdot5 = 10$. Multiplying these values of $b$ and $c$ by $10$, we get $b'=10$, $c'=50$, $A'=10\cdot20\cdot50 = 100^2 = (2^2\cdot5^2)^2$, which yields the $n=2$ case. This completes the construction.

Solution 2

We proceed via induction on n. Notice that we prove instead a stronger result: there exists a parabolic triangle with area $(2^nm)^2$ with two of the vertices sharing the same ordinate (y-coordinate).

Base case: If n = 0, consider the parabolic triangle ABC with A(0, 0), B(1, 1), C(-1, 1) that has area 1/2 * 1 * 2 = 1, so that n = 0 and m = 1. If n = 1, let ABC = A(5, 25), B(4, 16), C(-4, 16). Because ABC has area 1/2 * 8 * 9 = 36, we set n = 1 and m = 3. If n = 2, consider the triangle formed by A(21, 441), B(3, 9), C(-3, 9). It is parabolic and has area 1/2 * 6 * 432 = 1296 = $36^2$, so n = 2 and m = 9.

Inductive step: If n = k produces parabolic triangle ABC with A(a, $a^2$), B(b, $b^2$), and C(-b, $b^2$), consider A'B'C' with vertices A(4a, $16a^2$), B(4b, $16b^2$), and C(-4b, $16b^2$). If ABC has area $(2^km)^2$, then A'B'C' has area $(2^{k+3}m)^2$, which is easily verified using the 1/2 * base * height formula for triangle area. This completes the inductive step for k -> k+3.

Hence, for every nonnegative integer n, there exists an odd m and a parabolic triangle with area $(2^nm)^2$ with two vertices sharing the same ordinate. The problem statement is a direct result of this result.

See Also

2010 USAJMO (Problems • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6
All USAJMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2010_USAJMO_Problems/Problem_4&oldid=61449"