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Difference between revisions of "2010 USAJMO Problems/Problem 4"
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==Solution 2==
 
==Solution 2==
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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).
+
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 y-coordinate.
−
 
 
 
Base case:
 
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 <math>n = 0</math>, consider the parabolic triangle <math>ABC</math> with <math>A(0, 0), B(1, 1), C(-1, 1)</math> that has area <math>1/2 \cdot 1 \cdot 2 = 1</math>, so that <math>n = 0</math> and <math>m = 1</math>.
−
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 <math>n = 1</math>, let <math>ABC = A(5, 25), B(4, 16), C(-4, 16)</math>. Because <math>ABC</math> has area <math>1/2 \cdot 8 \cdot 9 = 36</math>, we set <math>n = 1</math> and <math>m = 3</math>.
−
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.
+
If <math>n = 2</math>, consider the triangle formed by <math>A(21, 441), B(3, 9), C(-3, 9)</math>. It is parabolic and has area <math>1/2 \cdot 6 \cdot 432 = 1296 = </math>36^2<math>, so </math>n = 2<math> and </math>m = 9<math>.
  
 
Inductive step:
 
Inductive step:
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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.
+
If n = k produces parabolic triangle ABC</math> with <math>A(a, a^2), B(b, b^2),</math> and <math>C(-b, b^2)</math>, consider <math>A'B'C'</math> 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)^2</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.
 +
 +
                                                                -MathGenius_
  
 
==Solution 3 (without induction)==
 
==Solution 3 (without induction)==

Revision as of 16:46, 25 February 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$ (Error compiling LaTeX. Unknown error_msg) 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. 

                                                               -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 $\frac{1}{2}bh=a(b^2-a^2)$. 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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