AoPS Academy
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 
be a circle and 
a given point in the plane. Each line through which intersects determines a chord of . Show that the midpoints of these chords lie on a circle.
be 
. is midpoint of the chord (specifically, diameter) through 
, so it is one of the points of interest. Let the midpoint of be 
. Let the midpoint of any one of the described chords be . Then 
because is the midpoint of one of the chords of , which has center . Then 
is right, meaning 
. Since is the center of an arbitrary chord, which contains when extended, all the midpoints of such chords lie on the circle with center and radius 
.
be a circle and a given point in the plane. Each line through which intersects determines a chord of . Show that the midpoints of these chords lie on a circle.
is midpoint of 
, we get that 
where is midpoint of such chords lie on a circle center at with radius 
. Could someone please elaborate on that?
is a part of a chord on circle 
which we can let be segment for 
on the circle. Since 
is isosceles and thus the median 
is also an altitude, from which the result follows.
![[asy]
size(200);
defaultpen(linewidth(0.8)+fontsize(12pt));
pair o=origin,p=(1.2,3.6),m=(-1.889,4.63),l=(4.289,2.57),po=(0,4),m2=(-3,4),l2=(3,4);
dot(po);
dot(p);
dot((0,0));
draw(m--l,dashed);
draw(m2--l2);
draw(o--p,dashed);
draw(po--o);
draw(circle(o,5));
label("$P_o$",po,SW);
label("$P_i$",p,SE);
label("$\omega$",o,SE);
label("$r_o$",0.5*po,W);
label("$r_i$",0.5*p,E);
label(scale(0.75)*"$\theta_i$",(0.26,1.65));
add(pathticks(po--m2,2,0.5,4,5));
add(pathticks(po--l2,2,0.5,4,5));
add(pathticks(p--m,3,0.5,4,5));
add(pathticks(p--l,3,0.5,4,5));
draw(anglemark(p,o,po,25));
draw(rightanglemark(po,p,o,12));
[/asy]](http://latex.artofproblemsolving.com/4/1/5/415c1abf15d262d6a3698085947e852b47f94be4.png)
(that fulfills the conditions of the problem), center , and 
is 
, the distance between and center is
(where 
is the distance between and the center of ), which is quite clearly a circle when graphed in polar coordinates.
and the center of the circle is the center of the locus?
. Consider an arbitrary chord with endpoints 
and 
on circle and let be the midpoint of . By definition, 
. Hence, we see that 
is a right triangle with hypotenuse . However, since this is an arbitrary chord, , all the chords passing through will form infinitely many right triangles with hypotenuse . Finally, to finish, we note that any triangle with hypotenuse has a circumcircle of diameter of . Since there are infinitely many right triangles with hypotenuse , this is the definition of a circle with diameter of and we are done. 
be the the centre of the circle and 
be the mid-points of the chords of passing through P.
If is the centre of a circle and is the mid-point of a chord XY of that circle, then :
and let it's mid-point be 
.
by the above mentioned lemma.
and similarly consider the mid-point of another chord 
. let this this mid-point be some 
. SO now by the lemme we have 
.
, it implies that 
is cyclic. Similarly we can go on taking distinct and proving that 
is cyclic. we still have 
and in common. Thus all the mid-points of the chords of lie on the circumcircle of 
. and the chord perpendicular to passing through . The midpoints of these two chords are respectively and . Now consider any other chord, , passing through and let be its midpoint. It is clear that 
. Thus, points 
lie on a circle with diameter . We can do the same thing for all other chords of passing through which finishes the problem. 
denote the centre of the circle is the midpoint itself becomes the midpoint
and lie on the circle be a chord of circle with on it and midpoint 
(perpendiculars from centre bisect the chord) too lies on the circle with diameter 
, we choose four points from the set of chords of that contain - 
. Let be the center of .
are cyclic.
lies on a circle. Which implies that every element from the set of chords of that contain must lie on 
.
, which we will denote as .
be the midpoints of the chords that go through point .
for all 
, so they all lie on a circle with diameter . 
be 
Let be the midpoint of an arbitrary chord 
It is well-known that 
(
by SSS). Therefore, 
implying that lies on the circle with diameter by intercepted arcs. Since the properties of apply to the midpoint of any chord passing through a point in the interior of 
all midpoints must lie on a circle with diameter 
so we are done.
as . Name the midpoints of two arbitrary cord that passes through as and which are not the diameter. Consider the quadrilateral 
. Since 
are midpoints we know that 
and 
. This means that is cyclic. Thus the circle with diameter contains all the midpoints of the chords that pass through .
be the midpoint of the chord determined by the line 
. Because intersects the circle , we know that the segment is a diameter of the circle. Because all diameters of a circle are congruent, it follows that all the midpoints of the chords determined by lines through are congruent. Therefore, all of these midpoints lie on a circle, which we will call 
. is indeed a circle, we must show that it satisfies the three properties of a circle: are congruent, it follows that all of the points on are equidistant from its center, which is the midpoint of the diameter defined by . are equidistant from its center, it follows that all points in the plane that are equidistant from its center are also on . are equidistant from its center, which is a fixed point, it follows that is the locus of all points in the plane that are equidistant from this fixed point. is a circle.
of any chord containing satisfies 
, where is the center of . Hence, all lie on the circle with diameter 
, which finishes. 
and getting nowhere ("is this some crazy homothety thing?"), before realizing that I could fix two of the points. The only points that one knows actually anything about are and (which are indeed midpoints), and substituting yields 
. At the time of solving I was in the car and consequently had to freedraw, which delayed my noticing: "wait, they're both right!"
with 
collinear, with the midpoint of being 
from congruent triangles. Hence, lies on the circle with diameter 
(where is the center of the circle) serving as the diameter.
. We see that, no matter where the chord is, we can draw the line segment 
to bisect the chord. Since lies on the chord, 
for any such . By the inscribed angle theorem, we can see all such form a circle with diameter , just like we had suspected. 
and let be the centre. Consider the diametre passing through . Draw a random chord 
containing but not . Name the midpoint of as . Notice that 
is isosceles and bisects and isosceles triangles have a cool property by which we obtain 
. Now by simply staring at the drawing, you realise that lies on the circle with diametre (due to Thales ). 
be the center of . We claim that all the midpoints lie on the circle with diameter . For any midpoint , 
, proving our claim, as desired.
for all midpoints. Therefore, it lies on the circle with diameter 
and the midpoints of all chords passing through be 
.
.
is cyclic 
(By Inscribed Angle Theorem).
lie on a circle with diameter .
, let and 
be two arbitrary fixed points on the conic, and let be a point in the plane. Let be a moving point on , let be the second intersection of 
with . Let 
, and let 
be the harmonic conjugate of in segment 
. Let be the pole of line 
in . Prove that moves on a conic through 
, and that 
is a harmonic quadrilateral on this conic." moves on a conic. Move with degree 
on , then by projecting through we get that moves with degree , and moves with degree 
. Consider the polar of in the fixed conic ; this line moves with degree by Plucker's formulas, and passes through the pole of line = by pole-polar duality. Therefore the intersection of the polar of with line 
(degree ) moves with degree 
, so moves on a fixed conic. = 
, and the tangent to in the locus of (let this tangent line be 
), we get that this conic passes through . through , since ![\[P(I,J;O,P) = (I, J, OP \cap IJ, \ell \cap IJ) = -1\]](http://latex.artofproblemsolving.com/d/f/c/dfcf3a454a0ba2263fea6a00e36bf7963c89c519.png)
(the final harmonicity follows by 
being the polar of ), we are done.
to the circle points, and we get that the midpoint of the circle chord moves on a conic 
such that 
, implying that 
, which is the circle with diameter .
be the centre of circle and , be the two feet of tangent WRT . THE CIRCLE in question is the circumcircle of 
.
is concyclic due to two right angles, giving that is the diameter of THE CIRCLE.
intersecting circle at and and the midpoint of is called 
. In we have 
. Now move your focus to THE CIRCLE. Basically, we have shown that 
, meaning 
(or 
) is concyclic. 
. be the midpoints of any two chords passing through . It's well known that lies on the perpendicular bisector of and , meaning that 
, so 
lie on a circle with diameter , which is true for any . 
![$ a,b\in[\frac{1}{2},1] $](http://latex.artofproblemsolving.com/0/d/2/0d24f742e1f7db553b3a8795c061c796f07a897f.png)
.
Let ![$ a,b\in[\frac{1}{2},2] $](http://latex.artofproblemsolving.com/a/8/d/a8d4c8205bac0e34129f582a8f255355eda12e0b.png)
.
is right. Let the midpoints of all the other chords through 
.
is right, so quadrilateral 
is cyclic. Since we have all the circles based on 
and 
We claim 
lies on the circle with diameter 
as 
by SSS. 
since this immediately implies that all such midpoints lie on the circle with diameter 
congruent to 
which immediately implies the result.
so ![[asy] pair p, a, b, m; a = dir(50); b = dir(150); p = 0.4 * a + 0.6 * b; m = (a+b)/2;
draw(circle(0, 1),blue); draw(circle(p/2, abs(p/2)),green);
draw(a--b, blue); draw(m--(0,0), blue);
dot("$P$", p, dir(100)); dot("$A$", a, dir(30)); dot("$B$", b, dir(150)); dot("$M$", m, dir(100)); dot("$O$", (0,0)); rightanglemark(p, m, (0,0));
[/asy]](http://latex.artofproblemsolving.com/5/a/e/5ae2f5b0119dda8705975bf1a68c1cfe4dc8fec5.png)
,
,
be the chords intersecting 
,
,
,
,
,
,
,
.
,
,
where 
.
be the midpoint of 
.