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satisfy that: 
and 
. Find the maximum: 
, such that 
Prove that :
equality cases : ?
with 
, 
and area 
. The circle 
with diameter 
meets 
at 
. Let the orthocenter of the triangle be 
, connect 
and extend to meet at 
such that 
and 
is the diameter of . Draw the circumcircle 
of 
, chord 
of is tangent to and it passes through , compute .
and 
Prove that
Let and 
Prove that
Let and 
Prove that
Let and 
Prove that
with 
. The points 
lie on 
respectively, such that 
. Extend 
to meet the circumcircle of at 
respectively. Let the circumcircles of 
meet at points 
. Extend 
to at 
, given 
, find ![$[ABC]$](http://latex.artofproblemsolving.com/d/3/3/d33cc80fa8f093e155c5be46d2e5d9da3d7e1ef5.png)
.
with 
, denote the incenter of the triangle as 
. Extend 
to meet the circumcircle of 
at 
, find the length of 
.
and 
. Prove that
Equality holds when 
Equality holds when 
Equality holds when 
" 
:
.
and 
, but because 
differs from 
, we have that 
, and denote 
.
, therefore 
,, and 
, so 
.
, we have that 
, and now just replacing it in 
, we obtain the desired result.
" 
:
and 
, again from the sin law we obtain that 
.
, therefore 
, so 
.
, we obtain 
, therefore replacing that in , we obtain that
, therefore 
, and thus the problem is solved.
is the reflection of 
through 
so 
and 
is not equal to so 
is a isosceles trapezium
meets 
at 
so 
are cyclic Thus 
is isosceles so 
be a convex quadrilateral with 
and let be the intersection point of it's diagonals.
. Remark. Easy and very nice problem !
. ( memorize it ! ):
. Thus, 
. 
.
. Thus, 
. 
a.s.o.
:
and 
are such that they have two equal sides and one eqaul angle, so 
or 
. In both cases, wanted equality follows almost directly.
doesn't work, because then 
, so the only case available is when the angles are suplementary.
and 
. We note that 
; hence it is sufficient to prove that the condition 
is equivalent to the condition 
.
if and only if 
. Since we know 
(or else triangles 
would be congruent and we would have 
) and 
, this later condition is equivalent to the condition 
, or 
, which gives us what we wanted.
, then we obtain 
, from the angle chasing, therefore this case doesn't work 
, because in the hypothesis we have that differs from .
and meet at P:
By the way, nice problem.
is equivalent to 
. Let 
be a point external to such that 
is a parallelogram. Since 
, 
is equivalent to 
bisects 
and meets the perpendicular bisector of 
at 
. This occurs if and only if 
is cyclic.
. Now note that since triangles and 
are isosceles,![\[\angle{PDB}=\angle{DAC}+\angle{ADB}=\angle{DBC}+\angle{ACB}=\frac{x}{2}\]](http://latex.artofproblemsolving.com/2/4/3/243bf0e4bc0a9f0e9ec83eaca95f376bec6ba779.png)
is cyclic if and only if 
or 
.
to such that 
. Angle chase to get the relationship between triangles 
and 
is SSA, so either 
or 
. The former gives , which is a contradiction. The latter gives the required. to such that 
. Again, angle chase to get that and 
which gives the required.
have same radii and hence if 
then either 
or their sum is 
.One gives contradiction to 
while other solves the problem 
So 
) and (
)
from here two cases exist.
which means Diagonals are equal contradiction.
Done!!!
and which means Diagonals are equal contradiction. 
or 
. From isosceles triangle 
and sine law:
and sine law:
and 
or 
and
and 
, 
, 
, 
. From Stewart's theorem in 
and 
, we have 
. Rearranging gives 
. Since 
by the condition of the problem, we obtain 
. be the reflection of across the midpoint of , and let be the reflection of 
across the midpoint of 
. Then, 
, 
, 
, and 
. Also, 
and 
. Therefore, 
and 
are equilateral triangles.
, we have 
. 
be the intersection of and 
, with 
. If we let 
, then 
, 
, and 
. on ![$[AE]$](http://latex.artofproblemsolving.com/0/e/e/0ee81ab9421e97824f7785dc2c44c0f5b04a03b1.png)
such that 
, we have as an equilateral triangle, and by similarity or symmetry, 
. Thus, 
. Similarly, taking on ![$[DE]$](http://latex.artofproblemsolving.com/4/f/5/4f55b2be1d3d9963afec61b4973bfecc6141b1ff.png)
gives us 
, we obtain the desired result.
![$ a,b\in[\frac{1}{2},1] $](http://latex.artofproblemsolving.com/0/d/2/0d24f742e1f7db553b3a8795c061c796f07a897f.png)
.
Let ![$ a,b\in[\frac{1}{2},2] $](http://latex.artofproblemsolving.com/a/8/d/a8d4c8205bac0e34129f582a8f255355eda12e0b.png)
.
.
.
and 
,
q.e.d.
is not equal with 60, we find # 