be the triangle and 
is arbitrary point. 
are nine-point centers of 
.
are concurrent.
board, a nonnegative real number is written in such a way that the sum of the numbers in each row is equal to 
,
to be the largest value in row 
,
.
to be the largest value in column 
.
?
which satisfy:
.
,
such that: 
and 
Where ![$ k\in(0,\frac{365(2-\sqrt2)}{54}]. $](http://latex.artofproblemsolving.com/b/5/a/b5a7af8a975aec3b431c2764c76d2f4cfda6c699.png)
Where 
Where 
Y by
In a triangle 
, 
the point 
is the midpoint of 
and 
is a point on the side such that 
. The cotangents of the angles 
, 
, and 
are positive integers 
respectively.
(a) Show that the cotangent of the angle
is also an integer and equals 
.
(b) Show that there are infinitely many possible triples
, some of which consisting of Fibonacci numbers.
, 
the point 
is the midpoint of 
and 
is a point on the side such that 
. The cotangents of the angles 
, 
, and 
are positive integers 
respectively.(a) Show that the cotangent of the angle
is also an integer and equals 
.(b) Show that there are infinitely many possible triples
, some of which consisting of Fibonacci numbers.Attachments:
This post has been edited 2 times. Last edited by Stear14, Jun 1, 2025, 6:41 PM
Reason: formatting
Reason: formatting
,
,
,
i
i
and the vertex 
i
.
and 
.
and 
is given by![\[
\cot \theta = \pm \frac{u_1v_1+u_2v_2}{u_1v_2-u_2v_1}
\]](http://latex.artofproblemsolving.com/1/9/a/19ac7a1663889672c0e51a4d21253c246c6e4a0e.png)
(the dot product divided by the area of the parallelogram spanned by these vectors). Using this formula, we obtain:![\[
a = \frac{x^2+y^2+3y}{3x}, \quad
b = \frac{x^2+y^2+7y+12}{x}, \quad
c= \frac{x^2+y^2+10y+24}{2x}
\]](http://latex.artofproblemsolving.com/f/6/a/f6a6259ce387551f3bf7f4d9257d72ddd369d15c.png)
After simple manipulations, we arrive at the following principal identity:![\[
ab+bc+ac=b^2+2 \hspace{1cm} (*)
\]](http://latex.artofproblemsolving.com/6/2/4/62446e349534579debc1f416e8cc2d1fd241f35f.png)
If we can find a solution to this equation in positive integers, then the coordinates 
can be found using the following formulae:![\[
x = \frac{12}{9a-7b+8c}, \quad
y=\frac{12(2b-3a-2c)}{9a-7b+8c}=x(2b-3a-2c)
\]](http://latex.artofproblemsolving.com/6/5/d/65d47ec4421395c444c79c5bb09c36df87937ffa.png)
is given by the formula![\[
\cot (\alpha+\beta+\gamma) =
\frac{\cot\alpha\cot\beta\cot\gamma-\cot\alpha-\cot\beta-\cot\gamma}{
\cot\alpha\cot\beta+\cot\alpha\cot\gamma+\cot\beta\cot\gamma-1}=
\frac{abc-a-b-c}{ab+ac+bc-1}\ ,
\]](http://latex.artofproblemsolving.com/c/6/2/c62dd39a109acd8d3c76c2921facd9c8c9ddf837.png)
which, using (*) simplifies to:![\[
\cot(\alpha+\beta+\gamma) = b-a-c,
\]](http://latex.artofproblemsolving.com/4/6/2/462bf072be68ff53d0ab61dcbbb5d1c54ff99200.png)
solving part (a) of the problem.
-![\[
F_n = \frac{\phi^n - (-\phi)^{-n}}{\sqrt{5}}\ ,
\]](http://latex.artofproblemsolving.com/a/d/d/add6db2f520b3cbf5417e249347c6a9102565d77.png)
it is easy to check that the principal identity ![\[
(a,b,c)=(F_{2n-3}, F_{2n}, F_{2n-1})\ .
\]](http://latex.artofproblemsolving.com/8/3/4/834941c5bbaf30908f0112d19edcc2b766ff2c54.png)
In this case we have:![\[
\cot(\alpha+\beta+\gamma)=F_{2n}-F_{2n-1}-F_{2n-3}=F_{2n-4}\ .
\]](http://latex.artofproblemsolving.com/c/5/b/c5b0c4754634371eb8118235f5355afcdbfcf548.png)
Another infinite series of solutions is given by the triples![\[
(a,b,c)=(L_{2n-2}, L_{2n}, F_{2n}), \quad
\cot(\alpha+\beta+\gamma)=b-a-c=F_{2n-2}\ .
\]](http://latex.artofproblemsolving.com/1/5/c/15c199636157329cb43a5ae2e34ab8640997acdc.png)
where 
is the 
.
there are at least two different solutions of the principal identit![\[
ac=bk+2 \quad {\rm and} \quad a+c=b-k
\]](http://latex.artofproblemsolving.com/4/e/7/4e793c4b2c4586149d4df5edd559e19669d832fc.png)
It has integer solutions when the discriminant of the corresponding quadratic equation is a perfect square: 
.
,![\[
4(8k^2+m^2+8)=p^2=(2q)^2\ .
\]](http://latex.artofproblemsolving.com/e/2/b/e2bf250cef0bf8d2ff5960e776c69e1849e62a85.png)
Therefore, we just need to find integer numbers, 
and 
such that 
.![\[
\{ q-m = 2, \quad q+m=4k^2+4\}\quad{\rm and}\quad
\{ q-m = 4, \quad q+m=2k^2+2\}
\]](http://latex.artofproblemsolving.com/8/c/8/8c800207195910f04b5b20ec5a93615c3a76651e.png)
we find two explicit solutions:![\[
(q,m)=(2k^2+3, 2k^2+1)\quad{\rm and} \quad
(q,m)=(k^2+3, k^2-1)
\]](http://latex.artofproblemsolving.com/3/1/c/31c18f0a92bb119a27eb3371e24e2409ca10ceac.png)
from which we can recover two different triples 
.
