be distinct natural numbers each of which has exactly 
positive integer factors. For instance, 
has exactly 
.
divides any of the 
'
be an integer. Prove that 
can be written as a polynomial with integer coefficients in the variables 
,
and 
.
,
are real numbers such that 
is an integral multiple of 
.
,
for all positive integers 
.
be the orthocenter of an acute-angled triangle 
.
centered at the midpoint of 
and passing through 
and 
.
,
,
and 
.
with real coefficient. It is known that 
and 
.
can't have 
distinct real roots.
Y by Adventure10, Mango247
Let 
be a right triangle with 
The bisectors of the angles 
and 
meet each other in 
and meet the sides 
and 
in 
and 
, respectively. Prove that 
where 
is the area function.
be a right triangle with 
The bisectors of the angles 
and 
meet each other in 
and meet the sides 
and 
in 
and 
, respectively. Prove that 
where 
is the area function. [asy]
import graph; size(200); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttqqcc = rgb(0.2,0,0.8); pen qqwuqq = rgb(0,0.39,0); pen xdxdff = rgb(0.49,0.49,1); pen fftttt = rgb(1,0.2,0.2); pen ccccff = rgb(0.8,0.8,1);
draw((1.89,4.08)--(1.89,4.55)--(1.42,4.55)--(1.42,4.08)--cycle,qqwuqq); draw((1.42,4.08)--(7.42,4.1),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(1.42,4.08),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(7.42,4.1),ttqqcc+linewidth(1.6pt)); draw((1.4,10.08)--(4,4.09),fftttt+linewidth(1.2pt)); draw((7.42,4.1)--(1.41,6.24),fftttt+linewidth(1.2pt)); draw((1.41,6.24)--(4,4.09),ccccff+linetype("5pt 5pt"));
dot((1.42,4.08),ds); label("$A$", (1.1,3.66),NE*lsf); dot((7.42,4.1),ds); label("$B$", (7.15,3.75),NE*lsf); dot((1.4,10.08),ds); label("$C$", (1.49,10.22),NE*lsf); dot((4,4.09),ds); label("$E$", (3.96,3.46),NE*lsf); dot((1.41,6.24),ds); label("$D$", (0.9,6.17),NE*lsf); dot((3.37,5.54),ds); label("$I$", (3.45,5.69),NE*lsf); clip((-6.47,-7.49)--(-6.47,11.47)--(16.06,11.47)--(16.06,-7.49)--cycle); [/asy]
![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
denote the area of 
.
be the inradius. Then ![$[BIC]= \frac{1}{2}(r)(BC)$](http://latex.artofproblemsolving.com/1/c/d/1cda9815afb605f3dc4c142bc1ebd8581bea3c68.png)
.![$[ABC]= \frac{1}{2}(AB)(AC)=\frac{1}{2}(r)(AB+BC+CA)$](http://latex.artofproblemsolving.com/b/5/c/b5ce915b65aedb65ee4cd1368387e1841d63a841.png)
,
.![$[BIC]=\frac{1}{2}\frac{(AB)(BC)(CA)}{AB+BC+CA}$](http://latex.artofproblemsolving.com/f/5/9/f59822b8c74fe2018aacf29eac76b650206848a1.png)
![$[BCDE]=[ABC]-[ADE]$](http://latex.artofproblemsolving.com/5/1/9/51903f3185fe9921c70d2afd1ca614bdbe0e23a5.png)
and 
![$[BCDE]=[ABC]-[ADE]= \frac{1}{2}(AB)(AC)- \frac{1}{2}\frac{(AB^2)(AC^2)}{(AB+BC)(AC+BC)}$](http://latex.artofproblemsolving.com/b/6/7/b672ffd7b304e296a6762e1bd284f9ad771666ce.png)
![$[BCDE]=[BIC] \iff AB^2+AC^2=BC^2$](http://latex.artofproblemsolving.com/3/c/6/3c63e6bcea8d78a59cb73630a1c0d409c54f8496.png)
,
a
on 
as 
and 
.
and 
.
.
and 
.
.
