Nine points circle theorem
by Nikpour, Apr 25, 2013, 2:06 PM
In this post, I will explore many results regarding the altitudes, the medians, the orthocenter and the centroid of a triangle
.
I will denote by
the feet of the altitudes with respect to
and
the midpoints of sides
. Also let
and
denote the orthocenter and centroid of
.
Let's start with a well-known result:
1:Nine-point circle: The points
are all concyclic.
Proof: Since
is right and
is the midpoint of
, we have that
is isosceles, thus
. Therefore,
are concyclic, and similarly all six points lie on a circle.
Remark. The midpoints of
are also in this circle, thus the name nine-point circle. The proof is analogous angle-chasing.
2:Prove that the ray
intersects the circumcircle of
at a point
such that
is parallel to
.
![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -2.55, xmax = 13.48, ymin = -3.63, ymax = 4.32; /* image dimensions */
pen ccwwqq = rgb(0.8,0.4,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); pen ffcctt = rgb(1,0.8,0.2); pen cccccc = rgb(0.8,0.8,0.8);
draw((2.24,3.52)--(0.66,-1.34)--(7.36,-1.36)--cycle, ccwwqq);
/* draw figures */
draw((2.24,3.52)--(0.66,-1.34), ccwwqq);
draw((0.66,-1.34)--(7.36,-1.36), ccwwqq);
draw((7.36,-1.36)--(2.24,3.52), ccwwqq);
draw(circle((4.01,0.26), 3.72), ttttff);
draw(circle((3.12,0.28), 1.86), ttttff);
draw((4.8,1.08)--(4.01,-1.35), ttzzqq);
draw((4.01,-1.35)--(1.45,1.09), ttzzqq);
draw((1.45,1.09)--(4.8,1.08), ttzzqq);
draw((2.24,3.52)--(5.81,3.51), red);
draw((2.24,3.52)--(2.23,-1.34), ffcctt);
draw((4.01,-1.35)--(2.24,3.52), cccccc);
draw((1.45,1.09)--(7.36,-1.36), cccccc);
draw((4.8,1.08)--(0.66,-1.34), cccccc);
draw((5.81,3.51)--(1.43,-2.42), linetype("4 4") + red);
draw((4.02,1.08)--(4.01,-1.35));
/* dots and labels */
dot((2.24,3.52),dotstyle);
label("$A$", (2.28,3.59), NE * labelscalefactor);
dot((0.66,-1.34),dotstyle);
label("$B$", (0.55,-1.54), NW * 2);
dot((7.36,-1.36),dotstyle);
label("$C$", (7.43,-1.58), NE * labelscalefactor);
dot((4.8,1.08),dotstyle);
label("$M_B$", (4.97,0.96), NE * labelscalefactor);
dot((1.45,1.09),dotstyle);
label("$M_C$", (1.18,1.01), W *0.5);
dot((4.01,-1.35),dotstyle);
label("$M_A$", (4,-1.53), S);
dot((3.42,0.27),dotstyle);
label("$G$", (3.23,0.36), NE * labelscalefactor);
dot((2.23,-1.34),dotstyle);
label("$H_A$", (2.26,-1.55), S);
dot((1.43,-2.42),dotstyle);
label("$T_A$", (1.34,-2.19), NE * labelscalefactor);
dot((5.81,3.51),dotstyle);
label("$T'_A$", (5.78,3.29), NE*2);
dot((4.01,0.26),dotstyle);
label("$O$", (4.06,0.33), NE * labelscalefactor);
dot((4.02,1.08),dotstyle);
label("$H'_A$", (3.76,1.2), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](//latex.artofproblemsolving.com/b/d/d/bdd24afb924a9283ce328f76376ac65336d90da0.png)
Proof: Let
denote the homothety centered at
which sends
to
. Then the image of
is the intersection of the nine-point circle of
and
, which is just
. But then
, so
.
3:Let
denote the circle passing through
,
tangent to the circumcircle
of
at a point
. Then
are collinear.
![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -3.59, xmax = 13.85, ymin = -4.12, ymax = 4.54; /* image dimensions */
pen ccwwqq = rgb(0.8,0.4,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); pen ffcctt = rgb(1,0.8,0.2); pen cccccc = rgb(0.8,0.8,0.8);
draw((2.14,3.44)--(0.66,-1.34)--(7.36,-1.36)--cycle, ccwwqq);
/* draw figures */
draw((2.14,3.44)--(0.66,-1.34), ccwwqq);
draw((0.66,-1.34)--(7.36,-1.36), ccwwqq);
draw((7.36,-1.36)--(2.14,3.44), ccwwqq);
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draw(circle((3.07,0.25), 1.85), ttttff);
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draw((4.01,-1.35)--(1.4,1.05), ttzzqq);
draw((2.14,3.44)--(5.91,3.43), red);
draw((2.14,3.44)--(2.12,-1.34), ffcctt);
draw((4.01,-1.35)--(2.14,3.44), cccccc);
draw((1.4,1.05)--(7.36,-1.36), cccccc);
draw((4.75,1.04)--(0.66,-1.34), cccccc);
draw((5.91,3.43)--(1.34,-2.33), linetype("4 4") + red);
draw((4.02,1.04)--(4.01,-1.35));
draw(circle((3.07,-0.67), 2.4), cccccc);
draw(circle((3.08,1.84), 1.85), cccccc);
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draw((-1.92,1.06)--(7.64,1.03), ttzzqq);
draw((5.91,3.43)--(0.4,1.05), red);
draw((5.91,3.43)--(7.64,1.03), red);
/* dots and labels */
dot((2.14,3.44),dotstyle);
label("$A$", (2.18,3.51), NE * labelscalefactor);
dot((0.66,-1.34),dotstyle);
label("$B$", (0.54,-1.56), W);
dot((7.36,-1.36),dotstyle);
label("$C$", (7.43,-1.59), NE * labelscalefactor);
dot((4.75,1.04),dotstyle);
label("$M_B$", (4.93,0.91), N*1.5);
dot((1.4,1.05),dotstyle);
label("$M_C$", (1.11,0.96), N);
dot((4.01,-1.35),dotstyle);
label("$M_A$", (4,-1.54), S);
dot((3.39,0.25),dotstyle);
label("$G$", (3.18,0.34), NE * labelscalefactor);
dot((2.12,-1.34),dotstyle);
label("$H_A$", (2.16,-1.57), S*0.25);
dot((1.34,-2.33),dotstyle);
label("$T_A$", (1.23,-2.07), S*4);
dot((5.91,3.43),dotstyle);
label("$T'_A$", (5.88,3.19), NE*2);
dot((4.01,0.24),dotstyle);
label("$O$", (4.06,0.32), NE * labelscalefactor);
dot((4.02,1.04),dotstyle);
label("$H'_A$", (3.74,1.16), N);
dot((-1.92,1.06),dotstyle);
label("$P_A$", (-1.98,1.22), NE * labelscalefactor);
dot((7.64,1.03),dotstyle);
label("$T_2$", (7.71,1.18), NE * labelscalefactor);
dot((0.4,1.05),dotstyle);
label("$T_1$", (0.17,1.19), W);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](//latex.artofproblemsolving.com/5/2/e/52e2a9d9d9e37ab276eaa607615515c51b5dfb44.png)
Proof: Let
denote the circumcircle of
, which is also tangent to
. Let
be the radical center of
, and let
be the intersections of
and
, as above. Then since
and
are tangent to
is a harmonic quadrilateral. Let
denote the second intersection of
and
, and
denote the image of
under the homothety
which sends
to
. Then
, so
and
is the midpoint of
. We have shown in (2) that
, so
is a harmonic pencil, which implies that
is a harmonic pencil, and
is a harmonic quadrilateral
, as desired.
Now, a simple but very useful lemma about the orthocenter:
4: Let
be the point diametrically opposite to
w.r.t.
. Prove that the ray
intersects
at
, and that
is also the midpoint of
.
![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 24.5, ymin = -8, ymax = 6.3; /* image dimensions */
pen ccwwqq = rgb(0.8,0.4,0); pen ffcctt = rgb(1,0.8,0.2);
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/* draw figures */
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draw((2,-3.04)--(10.76,-3.2), ccwwqq);
draw((10.76,-3.2)--(4,4.46), ccwwqq);
draw(circle((6.43,-0.21), 5.26));
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draw((6.85,1.24)--(2,-3.04), ffcctt);
draw((2,-3.04)--(8.87,-4.87));
draw((10.76,-3.2)--(8.87,-4.87));
draw((3.89,-1.37)--(8.87,-4.87));
draw((4,4.46)--(8.87,-4.87), red);
draw((3.86,-3.07)--(4,4.46), ffcctt);
/* dots and labels */
dot((4,4.46),dotstyle);
label("$A$", (3.82,4.58), NE * labelscalefactor);
dot((2,-3.04),dotstyle);
label("$B$", (1.7,-3.28), W*0.5);
dot((10.76,-3.2),dotstyle);
label("$C$", (10.94,-3.34), NE * labelscalefactor);
dot((2.54,-1.01),dotstyle);
label("$H_C$", (2.16,-0.88), W*0.5);
dot((6.85,1.24),dotstyle);
label("$H_B$", (6.92,1.36), NE * labelscalefactor);
dot((3.86,-3.07),dotstyle);
label("$H_A$", (3.98,-2.92), NE * labelscalefactor);
dot((6.38,-3.12),dotstyle);
label("$M_A$", (6.54,-2.98), NE * labelscalefactor);
dot((3.89,-1.37),dotstyle);
label("$H$", (3.98,-1.24), NE * labelscalefactor);
dot((8.87,-4.87),dotstyle);
label("$A'$", (9,-5.12), NE * labelscalefactor);
dot((6.43,-0.21),dotstyle);
label("$O$", (6.52,-0.08), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](//latex.artofproblemsolving.com/0/d/2/0d242164f81c969ee74d2ba56ab622a0d330d34e.png)
Proof: I claim that
is a parallelogram. Indeed,
since it is immediate that
. Analogously,
since we have
. Therefore,
is a parallelogram, thus
passes through
, and
is the midpoint of
.
5: (USA TSTST 2011 #4 restatement) Let rays
and
intersect the circumcircle
of
at
and
respectively. Then
are collinear, where
is defined as in (3).
![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -5.2, xmax = 15.2, ymin = -4.83, ymax = 5.3; /* image dimensions */
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draw((0.36,-0.65)--(-2.23,1.25), linetype("4 4") + red);
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draw((1.35,1.24)--(4,-3.32));
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label("$A$", (2.1,3.89), N * labelscalefactor);
dot((0.66,-1.34),dotstyle);
label("$B$", (0.53,-1.59), W * labelscalefactor);
dot((7.36,-1.36),dotstyle);
label("$C$", (7.44,-1.63), NE * labelscalefactor);
dot((4.7,1.23),dotstyle);
label("$M_B$", (4.92,1.07), E *0.2);
dot((1.35,1.24),dotstyle);
label("$M_C$", (0.88,1.09), N);
dot((4.01,-1.35),dotstyle);
label("$M_A$", (4,-1.57), S * labelscalefactor);
dot((-2.23,1.25),dotstyle);
label("$P_A$", (-2.31,1.44), NE * labelscalefactor);
dot((7.79,1.22),dotstyle);
label("$T_2$", (7.88,1.39), NE * labelscalefactor);
dot((0.25,1.24),dotstyle);
label("$T_1$", (-0.01,1.4), NE * labelscalefactor);
dot((2.03,0.07),dotstyle);
label("$H$", (2.01,0.18), NE * labelscalefactor);
dot((0.36,-0.65),dotstyle);
label("$B_1$", (0.1,-0.92));
dot((4,-3.32),dotstyle);
label("$C_1$", (3.9,-3.54), S * labelscalefactor);
dot((0.67,2.4),dotstyle);
label("$C'$", (0.51,2.52), N * labelscalefactor);
dot((7.37,2.38),dotstyle);
label("$B'$", (7.42,2.46), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](//latex.artofproblemsolving.com/2/1/b/21b59434ea4e0f1537b9ffd2b1a577ccc86f3078.png)
Proof: Let
and
denote the antipodes of
and
w.r.t.
. From (4) we have that
and
are the midpoints of
and
, thus
, so
, hence
is cyclic with circumcircle
. Let
be the circumcircle of
. Then
is the radical center of
, therefore
are collinear, as desired.
6: (APMO 2012 #4 restatement) Let
be an acute triangle with circumcircle
. Denote by
the intersection of the ray
with
and by
the second intersection of
and
. Prove that the quadrilateral
is harmonic.
First, I will prove a lemma about harmonic quadrilaterals:
Let
be a harmonic quadrilateral with circumcircle
and let
be any point in the plane different from
. Denote by
the second intersections of
and
. Then
is also a harmonic quadrilateral.
Proof of lemma: Using similar triangles, it's not hard to find that:
,
where
denotes the power of
w.r.t.
. Thus
is harmonic.
Main Proof: Let
denote the point diametrically opposite to
w.r.t.
, and
the reflection of
w.r.t.
. It is easy to prove that
, and since
is the midpoint of
, we have
is harmonic, so
is a harmonic quadrilateral. Applying the above lemma to
and taking
gives the desired result.
7: (OBM (Brazil MO) 2011 #5 restatement) Let
be an acute triangle with circumcircle
. Let the circumcircle of
intersect
at
. Prove that the angle bisectors of
and
intersect at a point on side
.
![[asy]
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pen dotstyle = black; /* point style */
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draw((10.8,-2.78)--(5.54,4.3), ccwwqq);
draw(circle((5.53,1.31), 2.99), zzzzzz);
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draw((10.8,-2.78)--(5.52,-3.86));
draw((5.52,-3.86)--(4.06,-2.76));
draw((4.06,-2.76)--(2.95,-0.2));
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draw((5.54,4.3)--(5.52,-3.86), ffcctt);
draw((9.34,-3.87)--(2.95,-0.2));
draw((4.34,-1.43)--(10.8,-2.78), ffcctt);
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dot((4.06,-2.76),dotstyle);
label("$B$", (4.14,-2.64), NE * labelscalefactor);
dot((10.8,-2.78),dotstyle);
label("$C$", (10.88,-2.66), NE * labelscalefactor);
dot((4.34,-1.43),dotstyle);
label("$H_C$", (4.42,-1.3), NE * labelscalefactor);
dot((8.39,0.46),dotstyle);
label("$H_B$", (8.48,0.58), NE * labelscalefactor);
dot((2.95,-0.2),dotstyle);
label("$F \equiv A_1$", (2.56,-0.18), S*3);
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label("$H$", (5.6,-1.56), NE * labelscalefactor);
dot((5.52,-3.86),dotstyle);
label("$H'$", (5.6,-3.74), NE * labelscalefactor);
dot((7.43,-2.77),dotstyle);
label("$M_A$", (7.52,-2.66), NE * labelscalefactor);
dot((5.52,-2.76),dotstyle);
label("$H_A$", (5.6,-2.64), NE * labelscalefactor);
dot((9.34,-3.87),dotstyle);
label("$A'$", (9.42,-3.74), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](//latex.artofproblemsolving.com/4/e/4/4e45d1987975c22da3f546bb4482b90ffd0ad36d.png)
Proof: From the angle bisector theorem, the problem statement is equivalent to proving that
. This looks a lot like a harmonic quadrilateral, but of course
. Fortunately, we can take the reflection
of
w.r.t.
, so that
and
. Now, it remains to prove that
is a harmonic quadrilateral. But in (6) we have shown that
is a harmonic quadrilateral, where
is the intersection of the ray
and
.
Thus, it suffices to prove
are concyclic. Notice that
is clearly in the circumcircle of
. But then
, hence
is also in the circumcircle of
, so
, as desired.
8: Let
be four concyclic points. Let
denote the Simson line of
w.r.t.
, and define
similarly. Let
denote the line connecting
to the orthocenter of
, and define
similarly. Prove that the eight lines
are all concurrent.
![[asy]
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pen dotstyle = black; /* point style */
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draw((1.7,3.6)--(1.68,0.5), wwwwww);
draw((2.09,4.74)--(0.64,0.5), ccwwqq);
draw((2.09,4.74)--(4.96,0.48), linewidth(1.6) + qqwwqq);
draw(circle((1.17,2.05), 1.64), red);
draw(circle((3.33,3.68), 1.64), red);
draw(circle((5.61,2.12), 1.76), red);
draw(shift((3.45,0.49))*xscale(2.81)*yscale(2.81)*arc((0,0),1,-0.2,179.8), red);
/* dots and labels */
dot((1.7,3.6),dotstyle);
label("$A$", (1.74,3.66), NE * labelscalefactor);
dot((0.64,0.5),dotstyle);
label("$B$", (0.68,0.56), NE * labelscalefactor);
dot((6.26,0.48),dotstyle);
label("$C$", (6.31,0.54), NE * labelscalefactor);
dot((4.97,3.76),dotstyle);
label("$D$", (5.01,3.82), NE * labelscalefactor);
dot((3.85,2.13),dotstyle);
label("$H_{DB}$", (3.9,2.19), NE * labelscalefactor);
dot((4.96,0.48),dotstyle);
label("$H_{DA}$", (5,0.55), NE * labelscalefactor);
dot((1.69,2.04),dotstyle);
label("$H_D$", (1.73,2.1), NE * labelscalefactor);
dot((1.68,0.5),dotstyle);
label("$H_{AD}$", (1.73,0.56), NE * labelscalefactor);
dot((2.42,3.1),dotstyle);
label("$H_{BD}$", (2.46,3.17), NE * labelscalefactor);
dot((1.22,2.2),dotstyle);
label("$H_{CD}$", (1.26,2.26), NE * labelscalefactor);
dot((2.8,2.13),dotstyle);
label("$H_{AC}$", (2.84,2.2), NE * labelscalefactor);
dot((4.58,4.73),dotstyle);
label("$H_{AB}$", (4.63,4.79), NE * labelscalefactor);
dot((4.21,3.19),dotstyle);
label("$H_{CA}$", (4.25,3.25), NE * labelscalefactor);
dot((5.5,2.41),dotstyle);
label("$H_{BA}$", (5.54,2.48), NE * labelscalefactor);
dot((4.96,2.2),dotstyle);
label("$H_{A}$", (5.01,2.27), NE * labelscalefactor);
dot((0.49,3.53),dotstyle);
label("$H_{BC}$", (0.53,3.6), NE * labelscalefactor);
dot((6.09,3.82),dotstyle);
label("$H_{CB}$", (6.13,3.88), NE * labelscalefactor);
dot((6.02,5.3),dotstyle);
label("$H_{B}$", (6.06,5.36), NE * labelscalefactor);
dot((0.4,5.32),dotstyle);
label("$H_{C}$", (0.44,5.38), NE * labelscalefactor);
dot((3.33,2.9),dotstyle);
dot((2.09,4.74),dotstyle);
label("$H_{DC}$", (2.13,4.81), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](//latex.artofproblemsolving.com/3/8/0/380d201c833264295d73787ff96eda5104acf2db.png)
Proof: Denote by
the foot of the perpendicular from
to the segment
(the one which does not contain
), and similarly define the
other points of this type, as above. Denote by
the orthocenter of
, and define
similarly. Observe that the red circles illustrated above with diameters
contain each four other points.
From Pascal's Theorem on
, we get that
are concurrent. Analogously, Pascal's Theorem on
gives that
are concurrent. Similarly, all eight lines are concurrent, so done.

I will denote by







Let's start with a well-known result:
1:Nine-point circle: The points

Proof: Since






Remark. The midpoints of

2:Prove that the ray





![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -2.55, xmax = 13.48, ymin = -3.63, ymax = 4.32; /* image dimensions */
pen ccwwqq = rgb(0.8,0.4,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); pen ffcctt = rgb(1,0.8,0.2); pen cccccc = rgb(0.8,0.8,0.8);
draw((2.24,3.52)--(0.66,-1.34)--(7.36,-1.36)--cycle, ccwwqq);
/* draw figures */
draw((2.24,3.52)--(0.66,-1.34), ccwwqq);
draw((0.66,-1.34)--(7.36,-1.36), ccwwqq);
draw((7.36,-1.36)--(2.24,3.52), ccwwqq);
draw(circle((4.01,0.26), 3.72), ttttff);
draw(circle((3.12,0.28), 1.86), ttttff);
draw((4.8,1.08)--(4.01,-1.35), ttzzqq);
draw((4.01,-1.35)--(1.45,1.09), ttzzqq);
draw((1.45,1.09)--(4.8,1.08), ttzzqq);
draw((2.24,3.52)--(5.81,3.51), red);
draw((2.24,3.52)--(2.23,-1.34), ffcctt);
draw((4.01,-1.35)--(2.24,3.52), cccccc);
draw((1.45,1.09)--(7.36,-1.36), cccccc);
draw((4.8,1.08)--(0.66,-1.34), cccccc);
draw((5.81,3.51)--(1.43,-2.42), linetype("4 4") + red);
draw((4.02,1.08)--(4.01,-1.35));
/* dots and labels */
dot((2.24,3.52),dotstyle);
label("$A$", (2.28,3.59), NE * labelscalefactor);
dot((0.66,-1.34),dotstyle);
label("$B$", (0.55,-1.54), NW * 2);
dot((7.36,-1.36),dotstyle);
label("$C$", (7.43,-1.58), NE * labelscalefactor);
dot((4.8,1.08),dotstyle);
label("$M_B$", (4.97,0.96), NE * labelscalefactor);
dot((1.45,1.09),dotstyle);
label("$M_C$", (1.18,1.01), W *0.5);
dot((4.01,-1.35),dotstyle);
label("$M_A$", (4,-1.53), S);
dot((3.42,0.27),dotstyle);
label("$G$", (3.23,0.36), NE * labelscalefactor);
dot((2.23,-1.34),dotstyle);
label("$H_A$", (2.26,-1.55), S);
dot((1.43,-2.42),dotstyle);
label("$T_A$", (1.34,-2.19), NE * labelscalefactor);
dot((5.81,3.51),dotstyle);
label("$T'_A$", (5.78,3.29), NE*2);
dot((4.01,0.26),dotstyle);
label("$O$", (4.06,0.33), NE * labelscalefactor);
dot((4.02,1.08),dotstyle);
label("$H'_A$", (3.76,1.2), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](http://latex.artofproblemsolving.com/b/d/d/bdd24afb924a9283ce328f76376ac65336d90da0.png)
Proof: Let










3:Let







![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -3.59, xmax = 13.85, ymin = -4.12, ymax = 4.54; /* image dimensions */
pen ccwwqq = rgb(0.8,0.4,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); pen ffcctt = rgb(1,0.8,0.2); pen cccccc = rgb(0.8,0.8,0.8);
draw((2.14,3.44)--(0.66,-1.34)--(7.36,-1.36)--cycle, ccwwqq);
/* draw figures */
draw((2.14,3.44)--(0.66,-1.34), ccwwqq);
draw((0.66,-1.34)--(7.36,-1.36), ccwwqq);
draw((7.36,-1.36)--(2.14,3.44), ccwwqq);
draw(circle((4.01,0.24), 3.71), ttttff);
draw(circle((3.07,0.25), 1.85), ttttff);
draw((4.75,1.04)--(4.01,-1.35), ttzzqq);
draw((4.01,-1.35)--(1.4,1.05), ttzzqq);
draw((2.14,3.44)--(5.91,3.43), red);
draw((2.14,3.44)--(2.12,-1.34), ffcctt);
draw((4.01,-1.35)--(2.14,3.44), cccccc);
draw((1.4,1.05)--(7.36,-1.36), cccccc);
draw((4.75,1.04)--(0.66,-1.34), cccccc);
draw((5.91,3.43)--(1.34,-2.33), linetype("4 4") + red);
draw((4.02,1.04)--(4.01,-1.35));
draw(circle((3.07,-0.67), 2.4), cccccc);
draw(circle((3.08,1.84), 1.85), cccccc);
draw((2.14,3.44)--(-1.92,1.06));
draw((-1.92,1.06)--(1.34,-2.33));
draw((-1.92,1.06)--(7.64,1.03), ttzzqq);
draw((5.91,3.43)--(0.4,1.05), red);
draw((5.91,3.43)--(7.64,1.03), red);
/* dots and labels */
dot((2.14,3.44),dotstyle);
label("$A$", (2.18,3.51), NE * labelscalefactor);
dot((0.66,-1.34),dotstyle);
label("$B$", (0.54,-1.56), W);
dot((7.36,-1.36),dotstyle);
label("$C$", (7.43,-1.59), NE * labelscalefactor);
dot((4.75,1.04),dotstyle);
label("$M_B$", (4.93,0.91), N*1.5);
dot((1.4,1.05),dotstyle);
label("$M_C$", (1.11,0.96), N);
dot((4.01,-1.35),dotstyle);
label("$M_A$", (4,-1.54), S);
dot((3.39,0.25),dotstyle);
label("$G$", (3.18,0.34), NE * labelscalefactor);
dot((2.12,-1.34),dotstyle);
label("$H_A$", (2.16,-1.57), S*0.25);
dot((1.34,-2.33),dotstyle);
label("$T_A$", (1.23,-2.07), S*4);
dot((5.91,3.43),dotstyle);
label("$T'_A$", (5.88,3.19), NE*2);
dot((4.01,0.24),dotstyle);
label("$O$", (4.06,0.32), NE * labelscalefactor);
dot((4.02,1.04),dotstyle);
label("$H'_A$", (3.74,1.16), N);
dot((-1.92,1.06),dotstyle);
label("$P_A$", (-1.98,1.22), NE * labelscalefactor);
dot((7.64,1.03),dotstyle);
label("$T_2$", (7.71,1.18), NE * labelscalefactor);
dot((0.4,1.05),dotstyle);
label("$T_1$", (0.17,1.19), W);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](http://latex.artofproblemsolving.com/5/2/e/52e2a9d9d9e37ab276eaa607615515c51b5dfb44.png)
Proof: Let




























Now, a simple but very useful lemma about the orthocenter:
4: Let








![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 24.5, ymin = -8, ymax = 6.3; /* image dimensions */
pen ccwwqq = rgb(0.8,0.4,0); pen ffcctt = rgb(1,0.8,0.2);
draw((4,4.46)--(2,-3.04)--(10.76,-3.2)--cycle, ccwwqq);
/* draw figures */
draw((4,4.46)--(2,-3.04), ccwwqq);
draw((2,-3.04)--(10.76,-3.2), ccwwqq);
draw((10.76,-3.2)--(4,4.46), ccwwqq);
draw(circle((6.43,-0.21), 5.26));
draw((2.54,-1.01)--(10.76,-3.2), ffcctt);
draw((6.85,1.24)--(2,-3.04), ffcctt);
draw((2,-3.04)--(8.87,-4.87));
draw((10.76,-3.2)--(8.87,-4.87));
draw((3.89,-1.37)--(8.87,-4.87));
draw((4,4.46)--(8.87,-4.87), red);
draw((3.86,-3.07)--(4,4.46), ffcctt);
/* dots and labels */
dot((4,4.46),dotstyle);
label("$A$", (3.82,4.58), NE * labelscalefactor);
dot((2,-3.04),dotstyle);
label("$B$", (1.7,-3.28), W*0.5);
dot((10.76,-3.2),dotstyle);
label("$C$", (10.94,-3.34), NE * labelscalefactor);
dot((2.54,-1.01),dotstyle);
label("$H_C$", (2.16,-0.88), W*0.5);
dot((6.85,1.24),dotstyle);
label("$H_B$", (6.92,1.36), NE * labelscalefactor);
dot((3.86,-3.07),dotstyle);
label("$H_A$", (3.98,-2.92), NE * labelscalefactor);
dot((6.38,-3.12),dotstyle);
label("$M_A$", (6.54,-2.98), NE * labelscalefactor);
dot((3.89,-1.37),dotstyle);
label("$H$", (3.98,-1.24), NE * labelscalefactor);
dot((8.87,-4.87),dotstyle);
label("$A'$", (9,-5.12), NE * labelscalefactor);
dot((6.43,-0.21),dotstyle);
label("$O$", (6.52,-0.08), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](http://latex.artofproblemsolving.com/0/d/2/0d242164f81c969ee74d2ba56ab622a0d330d34e.png)
Proof: I claim that










5: (USA TSTST 2011 #4 restatement) Let rays








![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -5.2, xmax = 15.2, ymin = -4.83, ymax = 5.3; /* image dimensions */
pen ccwwqq = rgb(0.8,0.4,0); pen ttttff = rgb(0.2,0.2,1); pen ttzzqq = rgb(0.2,0.6,0); pen cccccc = rgb(0.8,0.8,0.8);
draw((2.04,3.81)--(0.66,-1.34)--(7.36,-1.36)--cycle, ccwwqq);
/* draw figures */
draw((2.04,3.81)--(0.66,-1.34), ccwwqq);
draw((0.66,-1.34)--(7.36,-1.36), ccwwqq);
draw((7.36,-1.36)--(2.04,3.81), ccwwqq);
draw(circle((4.02,0.52), 3.84), ttttff);
draw(circle((3.02,0.3), 1.92), ttttff);
draw((4.7,1.23)--(4.01,-1.35), ttzzqq);
draw((4.01,-1.35)--(1.35,1.24), ttzzqq);
draw(circle((3.03,2.17), 1.92), cccccc);
draw((2.04,3.81)--(-2.23,1.25));
draw((-2.23,1.25)--(7.79,1.22), ttzzqq);
draw(circle((3.02,-0.84), 2.66));
draw((4.7,1.23)--(7.37,2.38), linetype("4 4"));
draw((1.35,1.24)--(0.67,2.4), linetype("4 4"));
draw((0.36,-0.65)--(4,-3.32));
draw((0.36,-0.65)--(-2.23,1.25), linetype("4 4") + red);
draw((0.67,2.4)--(7.37,2.38), ttzzqq);
draw((0.36,-0.65)--(4.7,1.23));
draw((1.35,1.24)--(4,-3.32));
/* dots and labels */
dot((2.04,3.81),dotstyle);
label("$A$", (2.1,3.89), N * labelscalefactor);
dot((0.66,-1.34),dotstyle);
label("$B$", (0.53,-1.59), W * labelscalefactor);
dot((7.36,-1.36),dotstyle);
label("$C$", (7.44,-1.63), NE * labelscalefactor);
dot((4.7,1.23),dotstyle);
label("$M_B$", (4.92,1.07), E *0.2);
dot((1.35,1.24),dotstyle);
label("$M_C$", (0.88,1.09), N);
dot((4.01,-1.35),dotstyle);
label("$M_A$", (4,-1.57), S * labelscalefactor);
dot((-2.23,1.25),dotstyle);
label("$P_A$", (-2.31,1.44), NE * labelscalefactor);
dot((7.79,1.22),dotstyle);
label("$T_2$", (7.88,1.39), NE * labelscalefactor);
dot((0.25,1.24),dotstyle);
label("$T_1$", (-0.01,1.4), NE * labelscalefactor);
dot((2.03,0.07),dotstyle);
label("$H$", (2.01,0.18), NE * labelscalefactor);
dot((0.36,-0.65),dotstyle);
label("$B_1$", (0.1,-0.92));
dot((4,-3.32),dotstyle);
label("$C_1$", (3.9,-3.54), S * labelscalefactor);
dot((0.67,2.4),dotstyle);
label("$C'$", (0.51,2.52), N * labelscalefactor);
dot((7.37,2.38),dotstyle);
label("$B'$", (7.42,2.46), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](http://latex.artofproblemsolving.com/2/1/b/21b59434ea4e0f1537b9ffd2b1a577ccc86f3078.png)
Proof: Let


















6: (APMO 2012 #4 restatement) Let









First, I will prove a lemma about harmonic quadrilaterals:










Proof of lemma: Using similar triangles, it's not hard to find that:

where




Main Proof: Let













7: (OBM (Brazil MO) 2011 #5 restatement) Let








![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -4.3, xmax = 24.5, ymin = -8, ymax = 6.3; /* image dimensions */
pen ccwwqq = rgb(0.8,0.4,0); pen zzzzzz = rgb(0.6,0.6,0.6); pen ffcctt = rgb(1,0.8,0.2);
draw((5.54,4.3)--(4.06,-2.76)--(10.8,-2.78)--cycle, ccwwqq);
/* draw figures */
draw(circle((7.44,0.22), 4.5), blue);
draw((5.54,4.3)--(4.06,-2.76), ccwwqq);
draw((4.06,-2.76)--(10.8,-2.78), ccwwqq);
draw((10.8,-2.78)--(5.54,4.3), ccwwqq);
draw(circle((5.53,1.31), 2.99), zzzzzz);
draw(circle((6.49,0.77), 3.66), zzzzzz);
draw((10.8,-2.78)--(5.52,-3.86));
draw((5.52,-3.86)--(4.06,-2.76));
draw((4.06,-2.76)--(2.95,-0.2));
draw((8.39,0.46)--(4.06,-2.76), ffcctt);
draw((5.54,4.3)--(5.52,-3.86), ffcctt);
draw((9.34,-3.87)--(2.95,-0.2));
draw((4.34,-1.43)--(10.8,-2.78), ffcctt);
draw((5.54,4.3)--(2.95,-0.2));
/* dots and labels */
dot((5.54,4.3),dotstyle);
label("$A$", (5.62,4.42), NE * labelscalefactor);
dot((4.06,-2.76),dotstyle);
label("$B$", (4.14,-2.64), NE * labelscalefactor);
dot((10.8,-2.78),dotstyle);
label("$C$", (10.88,-2.66), NE * labelscalefactor);
dot((4.34,-1.43),dotstyle);
label("$H_C$", (4.42,-1.3), NE * labelscalefactor);
dot((8.39,0.46),dotstyle);
label("$H_B$", (8.48,0.58), NE * labelscalefactor);
dot((2.95,-0.2),dotstyle);
label("$F \equiv A_1$", (2.56,-0.18), S*3);
dot((5.52,-1.67),dotstyle);
label("$H$", (5.6,-1.56), NE * labelscalefactor);
dot((5.52,-3.86),dotstyle);
label("$H'$", (5.6,-3.74), NE * labelscalefactor);
dot((7.43,-2.77),dotstyle);
label("$M_A$", (7.52,-2.66), NE * labelscalefactor);
dot((5.52,-2.76),dotstyle);
label("$H_A$", (5.6,-2.64), NE * labelscalefactor);
dot((9.34,-3.87),dotstyle);
label("$A'$", (9.42,-3.74), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](http://latex.artofproblemsolving.com/4/e/4/4e45d1987975c22da3f546bb4482b90ffd0ad36d.png)
Proof: From the angle bisector theorem, the problem statement is equivalent to proving that












Thus, it suffices to prove







8: Let










![[asy]
import graph; size(10cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -3.44, xmax = 11.65, ymin = -1.79, ymax = 5.71; /* image dimensions */
pen ccwwqq = rgb(0.8,0.4,0); pen wwwwww = rgb(0.4,0.4,0.4); pen ffcctt = rgb(1,0.8,0.2); pen qqwwqq = rgb(0,0.4,0);
/* draw figures */
draw(circle((3.45,1.27), 2.92), blue);
draw((1.7,3.6)--(6.26,0.48), ccwwqq);
draw((4.97,3.76)--(0.64,0.5), ccwwqq);
draw((6.02,5.3)--(6.26,0.48), wwwwww);
draw((0.4,5.32)--(0.64,0.5), wwwwww);
draw((0.4,5.32)--(2.8,2.13), wwwwww);
draw((0.64,0.5)--(2.42,3.1), wwwwww);
draw((4.21,3.19)--(6.26,0.48), wwwwww);
draw((0.4,5.32)--(4.97,3.76), wwwwww);
draw((5.5,2.41)--(0.64,0.5), wwwwww);
draw((6.02,5.3)--(1.7,3.6), wwwwww);
draw((6.02,5.3)--(3.85,2.13), wwwwww);
draw((4.97,3.76)--(4.96,0.48), wwwwww);
draw((6.26,0.48)--(1.22,2.2), wwwwww);
draw((0.49,3.53)--(6.09,3.82), ccwwqq);
draw((0.64,0.5)--(6.26,0.48), ccwwqq);
draw((6.26,0.48)--(4.58,4.73), ccwwqq);
draw((0.4,5.32)--(6.26,0.48), linewidth(1.6) + ffcctt);
draw((6.02,5.3)--(0.64,0.5), linewidth(1.6) + ffcctt);
draw((1.7,3.6)--(4.96,2.2), linewidth(1.6) + ffcctt);
draw((4.97,3.76)--(1.69,2.04), linewidth(1.6) + ffcctt);
draw((4.58,4.73)--(1.68,0.5), linewidth(1.6) + qqwwqq);
draw((0.49,3.53)--(5.5,2.41), linewidth(1.6) + qqwwqq);
draw((1.22,2.2)--(6.09,3.82), linewidth(1.6) + qqwwqq);
draw((1.7,3.6)--(1.68,0.5), wwwwww);
draw((2.09,4.74)--(0.64,0.5), ccwwqq);
draw((2.09,4.74)--(4.96,0.48), linewidth(1.6) + qqwwqq);
draw(circle((1.17,2.05), 1.64), red);
draw(circle((3.33,3.68), 1.64), red);
draw(circle((5.61,2.12), 1.76), red);
draw(shift((3.45,0.49))*xscale(2.81)*yscale(2.81)*arc((0,0),1,-0.2,179.8), red);
/* dots and labels */
dot((1.7,3.6),dotstyle);
label("$A$", (1.74,3.66), NE * labelscalefactor);
dot((0.64,0.5),dotstyle);
label("$B$", (0.68,0.56), NE * labelscalefactor);
dot((6.26,0.48),dotstyle);
label("$C$", (6.31,0.54), NE * labelscalefactor);
dot((4.97,3.76),dotstyle);
label("$D$", (5.01,3.82), NE * labelscalefactor);
dot((3.85,2.13),dotstyle);
label("$H_{DB}$", (3.9,2.19), NE * labelscalefactor);
dot((4.96,0.48),dotstyle);
label("$H_{DA}$", (5,0.55), NE * labelscalefactor);
dot((1.69,2.04),dotstyle);
label("$H_D$", (1.73,2.1), NE * labelscalefactor);
dot((1.68,0.5),dotstyle);
label("$H_{AD}$", (1.73,0.56), NE * labelscalefactor);
dot((2.42,3.1),dotstyle);
label("$H_{BD}$", (2.46,3.17), NE * labelscalefactor);
dot((1.22,2.2),dotstyle);
label("$H_{CD}$", (1.26,2.26), NE * labelscalefactor);
dot((2.8,2.13),dotstyle);
label("$H_{AC}$", (2.84,2.2), NE * labelscalefactor);
dot((4.58,4.73),dotstyle);
label("$H_{AB}$", (4.63,4.79), NE * labelscalefactor);
dot((4.21,3.19),dotstyle);
label("$H_{CA}$", (4.25,3.25), NE * labelscalefactor);
dot((5.5,2.41),dotstyle);
label("$H_{BA}$", (5.54,2.48), NE * labelscalefactor);
dot((4.96,2.2),dotstyle);
label("$H_{A}$", (5.01,2.27), NE * labelscalefactor);
dot((0.49,3.53),dotstyle);
label("$H_{BC}$", (0.53,3.6), NE * labelscalefactor);
dot((6.09,3.82),dotstyle);
label("$H_{CB}$", (6.13,3.88), NE * labelscalefactor);
dot((6.02,5.3),dotstyle);
label("$H_{B}$", (6.06,5.36), NE * labelscalefactor);
dot((0.4,5.32),dotstyle);
label("$H_{C}$", (0.44,5.38), NE * labelscalefactor);
dot((3.33,2.9),dotstyle);
dot((2.09,4.74),dotstyle);
label("$H_{DC}$", (2.13,4.81), NE * labelscalefactor);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]](http://latex.artofproblemsolving.com/3/8/0/380d201c833264295d73787ff96eda5104acf2db.png)
Proof: Denote by









From Pascal's Theorem on




This post has been edited 2 times. Last edited by Nikpour, Jul 9, 2013, 5:14 PM