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Difference between revisions of "2010 USAJMO Problems/Problem 4"
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<math>A'=10\cdot20\cdot50 = 100^2 = (2^2\cdot5^2)^2</math>,
 
<math>A'=10\cdot20\cdot50 = 100^2 = (2^2\cdot5^2)^2</math>,
 
which yields the <math>n=2</math> case. This completes the construction.
 
which yields the <math>n=2</math> 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 <math>2^nm</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 = 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 = 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.
 +
 +
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.
 +
 +
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.
  
 
== See Also ==
 
== See Also ==

Revision as of 23:13, 11 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 $(2^nm)^2$.

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 $2^nm$ 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 = 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 = 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.

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(2a, $4a^2$), B(2b, $4b^2$), and C(-2b, $4b^2$). If ABC has area $2^km$, then A'B'C' has area $2^{k+3}m$, 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 $2^nm$ 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

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