be the orthocenter of acute triangle 
,
be the foot of the altitude from 
to 
,
be the reflection of 
.
intersects line 
and 
.
.
"
such that 
is an integer for every positive integer 
Prove that
,
are points on sides 
and 
.
and 
,
side length of 
area of 
is a right angle, and 
,
.
is a right angle, 
,
.
and 
both contain a right angle and 
through AA postulate. Because they are similar, 
,
,
,
.
.
,
.
and 
Prove that
Find 
.
Taking the square root of the second equation, we get
has a right angle at 
,
and 
.
and 
are points
and 
,
,
.
.
,
is a right angle, using pythagorean theorem, we can get 
.
,
.
from point 
be the intersection of said altitude and 
.
both have a right angle and 
,
through AA Postulate. 
,
,
.
(CD = DE, as stated in the problem), 
,
(they were formed by an altitude, and are therefore both right angles), we can show that 
through HL Theorem. 
corresponds with 
,
.
,
and 
.
.
,
,
are positive integers such that 
.
.
to rearrange the terms and place the higher-degree terms at the front of the expression in order to have a more recognizable form.
![\begin{align*}
a^4+b^4+c^4-2a^2b^2-2a^2c^2-2b^2c^2 &=a^4-2a^2(b^2+c^2)+b^4-2b^2c^2+c^4 \\
&= a^4-2a^2(b^2+c^2)+(b^2+c^2)^2 - 4b^2c^2 \\
&= [a^2 - (b^2+c^2)]^2-(2bc)^2 \\
&= (a^2-b^2-c^2+2bc)(a^2-b^2-c^2-2bc)\\
&=[a^2-(b-c)^2][a^2-(b+c)^2] \\
&= (a+b-c)(a-b+c)(a+b+c)(a-b-c) = -63.
\end{align*}](http://latex.artofproblemsolving.com/0/c/4/0c430b7562ec7571fdf3e492939c354e0c3e7e20.png)
,
,
,
,![\[
a^2(a+2b)(a-2b)=-63.
\]](http://latex.artofproblemsolving.com/1/0/0/1006c2fabc10f934c25b6e6ae822a59ce2dd2712.png)
and 
are the only square factors of 
,
.
are 
and 
,
are all distinct positive integers, there are no solutions.
.
.
is an imaginary number. Thus, N must be equal to +2. We then have
and 
,
.
.
.
as x varies over the positive real numbers.
.
.
Y by clarkculus, Pengu14, Exponent11
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