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

m (→‎Solution 2)
m (→‎Solution 3 (without induction))
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Then, consider triangle with vertices <math>(a,a^2), (-a,a^2), (b,b^2)</math>, and set <math>a=2^{2n}</math> and <math>b=2^{4n-2}+1</math>.
 
Then, consider triangle with vertices <math>(a,a^2), (-a,a^2), (b,b^2)</math>, and set <math>a=2^{2n}</math> and <math>b=2^{4n-2}+1</math>.
−
The area of this triangle is <math>\frac{1}{2}bh=a(b^2-a^2)</math>.
+
The area of this triangle is <math>\frac{1}{2} \cdot base \cdot height=a(b^2-a^2)</math>.
 
We have that <math>b^2-a^2=2^{8n-4}+2^{4n-1}+1-2^{4n}=2^{8n-4}-2^{4n-1}+1=(2^{4n-2}-1)^2</math>
 
We have that <math>b^2-a^2=2^{8n-4}+2^{4n-1}+1-2^{4n}=2^{8n-4}-2^{4n-1}+1=(2^{4n-2}-1)^2</math>
 
We desire <math>A=a(b^2-a^2)=2^{2n}m^2</math>, or <math>2^{4n-1}-1=m</math>, and <math>m</math> is clearly always odd for positive <math>n</math>, completing the proof.
 
We desire <math>A=a(b^2-a^2)=2^{2n}m^2</math>, or <math>2^{4n-1}-1=m</math>, and <math>m</math> is clearly always odd for positive <math>n</math>, completing the proof.

Revision as of 15:37, 5 June 2020

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 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 $1/2 \cdot 1 \cdot 2 = 1$, 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 $1/2 \cdot 8 \cdot 9 = 36$, 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 $1/2 \cdot 6 \cdot 432 = 1296 = 36^2$, 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 $1/2 \cdot\text{base} \cdot \text{height}$ formula for triangle area. This completes the inductive step for $k \implies k+3$.

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. -MathGenius_

Solution 3 (without induction)

First, consider triangle with vertices $(0,0)$, $(1,1)$, $(-1,1)$. This has area $1$ so $n=0$ case is satisfied.

Then, consider triangle with vertices $(a,a^2), (-a,a^2), (b,b^2)$, and set $a=2^{2n}$ and $b=2^{4n-2}+1$. The area of this triangle is $\frac{1}{2} \cdot base \cdot height=a(b^2-a^2)$. We have that $b^2-a^2=2^{8n-4}+2^{4n-1}+1-2^{4n}=2^{8n-4}-2^{4n-1}+1=(2^{4n-2}-1)^2$ We desire $A=a(b^2-a^2)=2^{2n}m^2$, or $2^{4n-1}-1=m$, and $m$ is clearly always odd for positive $n$, completing the proof.

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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