be a convex quadrilateral inscribed in a circle and satisfying 
.
and 
such that 
and 
.
.
.
on 
and 
on 
and 
,
lies on the circumcircle of triangle 
be 
.
as the intersection of the perpendicular bisector of 
and 
;
.
,
,
.
concur at 
,
.
lie on the nine-point circle of 
,
lies on 
,
.
is the common perpendicular bisector of 
and 
,
.
.
,
,
,
,
be a point on the side 
and 
are the feet of the altitudes from 
is the circumcircle of 
,
be the point on 
is a diameter of 
the circumcircle of triangle 
,
be the point such that 
is a diameter of 
and 
be the second intersection of 
.
w.r.t 
is cyclic. 
,
are collinear by Reim's theorem; similarly, 
are collinear, and the result follows.
Y by Amir Hossein, icematrix2, HWenslawski, megarnie, bjump, ihatemath123, ImSh95, Adventure10, Mango247, Rounak_iitr
Triangle 
is inscribed in circle 
with 
, 
, and 
. The bisector of angle 
meets side 
at 
and circle at a second point 
. Let 
be the circle with diameter 
. Circles and meet at and a second point 
. Then 
, where m and n are relatively prime positive integers. Find 
.
is inscribed in circle 
with 
, 
, and 
. The bisector of angle 
meets side 
at 
and circle at a second point 
. Let 
be the circle with diameter 
. Circles and meet at and a second point 
. Then 
, where m and n are relatively prime positive integers. Find 
.This post has been edited 1 time. Last edited by ksun48, Mar 29, 2012, 4:46 PM
hit 
.
is right, so if 
is the center of 
.
and 
.
,
.
.
.
.
.
.
.
,
.
.
.
,
.
(squared), we have 
,
,
.
Also, 
and 
Thus 
is equilateral. Notice now that 
But 
so 
bisects 
Thus, 
By law of cosines on 
we find 
But by ptolemy on 
,
so 
so 
,
,
hits the y-axis. Either gives 
is clearly equilateral, the circumcenter of 
is at 
and the circumradius is 
.
and the length of 
so that circle has radius 
so 
being a solution means that they cancel out!)
is then:![$\frac{1}{19 \cdot 7}^2 \left[16^2(19-49)^2+(15\sqrt3)^2(19+49)^2\right]$](http://latex.artofproblemsolving.com/c/8/f/c8f385629213309ae3862d22b7c47ffeb041bd6e.png)
;
we have 
a diameter, where 
,
is cyclic; now![\[\measuredangle{FAD}=\measuredangle{FAE}=\measuredangle{FNE}=\measuredangle{DNM}=\measuredangle{DAM},\]](http://latex.artofproblemsolving.com/e/2/2/e223f340ab3b4f4df159d705bf1f0961b6fab5ad.png)
(to avoid directed angles, we can also note that 
,
is the reflection of 
over 
and 
,
and 
.![\[AF^2=\frac{b^2c^2}{AM^2}=\frac{4b^2c^2}{2b^2+2c^2-a^2}=\frac{2(3)^2(5)^2}{2(3^2+5^2)-7^2}=\frac{30^2}{2(34)-49}=\frac{900}{19},\]](http://latex.artofproblemsolving.com/0/3/4/03449f896916bb558ccbda91ba9f7560120fa615.png)
so we're done.
,
.![$[ABC] = \frac{15\sqrt{3}}{4}$](http://latex.artofproblemsolving.com/5/3/4/53475c1f2feac24f2a7dab8a94fa0efdf711d9bf.png)
to show the radius of circle 
and cevian 
to show 
.
to show 
.
.
to show 
with cevian 
to show 
and hence 
.
.
and 
lie on circle 
,
.
meet circle 
at 
.
is the diameter of circle 
and 
are right triangles. Furthermore, 
lie on circle 
.
.
and 
.
and 
and let their meeting point be 
.
,
is parallel to 
.
.
is similar to 
.
,
and 
.
.
is similar to 
.
.![\[ \cos\angle{DEF}=\cos\angle{GCM}=\frac{CM}{CG}=\frac{16\sqrt{19}}{133} \]](http://latex.artofproblemsolving.com/f/6/e/f6e989ebc6b7c502225234036a766c6fac3c8bf0.png)
![\[ AF^{2}=AE^{2}+EF^{2}-2(AE)(EF)\cos\angle{DEF}\]](http://latex.artofproblemsolving.com/3/8/c/38c8d7fb25a84e11c05a26b223c62f42ac44876c.png)
![\[ AF^{2}=8^{2}+(\frac{14\sqrt{19}}{19})^{2}-2(8)(\frac{14\sqrt{19}}{19})(\frac{16\sqrt{19}}{133})=\frac{900}{19} \]](http://latex.artofproblemsolving.com/7/8/8/788e3813b9842fb4c1b939b913c7a31bebd536eb.png)
intersect the circumcircle of 
.
bisects angle 
,
,
is a diameter. Thus 
is a right angle. It is easy to see that 
,
,
are cyclic. So their pairwise radical axes, 
are concurrent at a point, 
,
is right, so 
are collinear. As 
is the orthric triangle of 
,
, GA,GB,GF,GC is harmonic, and thus ABFC is harmonic, which says AF is symmedian.
.
,
.
inversion. We get that 
,
and since the circumcircle gets inverted to 
,
is the foot of the perpendicular of 
to 
,
.
.
is a harmonic quadrilateral. Furthermore, a quick Law of Cosines check confirms that 
.
.
and 
,
and 
.
and 
.
,
is right. Since 
followed by a reflection across the angle bisector of 
.
.
By symmedian properties we know that 
,
,
,
,
.
.
.
giving us that 
.
.
,
where 
is the radius of 
.
to find this.
,
,
,
by Ptolemy’s on 
.
.
.
.
and 
are similar, where 
so now 
yielding the answer as 
.
.
.
is the perpendicular bisector of 
,
.
be 
.
which can be shown easily. Then 
on triangle 
Let 
.![$[ABC] = \frac{15 \sqrt{3}}{4}$](http://latex.artofproblemsolving.com/0/4/2/042ee4f3cb7c1c6f4a74bfacbb4599f2f2553574.png)
,
and 
.
,
.
.)
,
passes through the arc midpoint of 
and note that this means 
and 
are cyclic where 
.
is cyclic we have 
.
so now we calculate 
.
and 
we have 
.
is 30-60-90 so 
.
.
.
,
.
.
.
so 
by the barycentric distance formula. Fortunately, the sides of this triangle are small integers and after a bit of computation, we find 
.
,
,
is equilateral and thus 
.
,
,
;
.
.
.
to get the value of 
.
.
.
is negative.
is isosceles, we have 
By the angle bisector theorem, we also have 
Note that 
.
.
,
(We get 
.
.
.
Now we know that 
,
,
,
.![\[\omega: \ \left(x-\frac{5}{2} \right)^2 + \left(y-\frac{11}{2\sqrt{3}} \right)^2 = \frac{49}{3},\]](http://latex.artofproblemsolving.com/0/4/6/046814bd1af8dcdddd4bd0de928f1e823de1d372.png)
![\[\overline{BC}: \ y = -\frac{3\sqrt{3}}{13}x +\frac{15\sqrt{3}}{13}\]](http://latex.artofproblemsolving.com/e/6/7/e6714c662f61578e31c54f5fcc277618e09cef5b.png)
![\[\overline{AE}: \ y = \sqrt{3}x. \]](http://latex.artofproblemsolving.com/c/f/7/cf76ca96ce69bc3487427a92bf4d6cf4843dc0ae.png)
and 
This means that 
.![\[\gamma: \ \left(x-\frac{79}{32} \right)^2 + \left(y-\frac{79\sqrt{3}}{32} \right)^2 = \frac{2401}{256}\]](http://latex.artofproblemsolving.com/d/f/4/df49a68db444a670a4a6486834ccc6d7ab474951.png)
![\[\implies F = \left(\frac{15}{19}, \frac{75\sqrt{3}}{19} \right).\]](http://latex.artofproblemsolving.com/9/f/1/9f1fe6ad734111fa41c7e3af9b0804e54515c0e6.png)
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