Difference between revisions of "2010 USAJMO Problems/Problem 4"

Revision as of 20:54, 14 April 2014

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.

@@ Line 47: / Line 47: @@
 ==Solution 2==
-We proceed via induction on n. Notice that we prove instead a stronger result: there exists a parabolic triangle with area <math>2^nm</math> with two of the vertices sharing the same ordinate (y-coordinate).
+We proceed via induction on n. Notice that we prove instead a stronger result: there exists a parabolic triangle with area <math>(2^nm)^2</math> 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 m = 1.
+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(1, 1), B(2, 4), C(-2, 4). Because ABC has area 1/2 * 3 * 4 = 6, we set n = 1 and m = 3.
+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(3, 9), B(4, 16), C(-4, 16). It is parabolic and has area 1/2 * 7 * 8 = 28, so n = 2 and m = 7.
+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 = <math>36^2</math>, so n = 2 and m = 9.
 Inductive step:
-If n = k produces parabolic triangle ABC with A(a, <math>a^2</math>), B(b, <math>b^2</math>), and C(-b, <math>b^2</math>), consider A'B'C' with vertices A(2a, <math>4a^2</math>), B(2b, <math>4b^2</math>), and C(-2b, <math>4b^2</math>). If ABC has area <math>2^km</math>, then A'B'C' has area <math>2^{k+3}m</math>, which is easily verified using the 1/2 * base * height formula for triangle area. This completes the inductive step for k -> k+3.
+If n = k produces parabolic triangle ABC with A(a, <math>a^2</math>), B(b, <math>b^2</math>), and C(-b, <math>b^2</math>), consider A'B'C' with vertices A(4a, <math>16a^2</math>), B(4b, <math>16b^2</math>), and C(-4b, <math>16b^2</math>). If ABC has area <math>(2^km)^2</math>, then A'B'C' has area <math>(2^{k+3}m)^2</math>, 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 <math>2^nm</math> with two vertices sharing the same ordinate. The problem statement is a direct result of this result.
+Hence, for every nonnegative integer n, there exists an odd m and a parabolic triangle with area <math>(2^nm)^2</math> 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

Page

Toolbox

Search