AoPS Academy
be a convex quadrilateral and let the diagonals 
and 
intersect at 
. Let 
be respectively the incentres of triangles 
. Let 
be respectively the excentres of triangles opposite . Show that lie on a circle if and only if lie on a circle.
be a triangle with centroid 
. Points 
and 
are chosen on rays 
and 
, respectively, such that![\[ \angle ABS=\angle ACR=180^\circ-\angle BGC.\]](http://latex.artofproblemsolving.com/0/a/d/0ad01e29194358ea0d3e10cb74b0769d1bd3a656.png)
Prove that 
.
have orthocenter 
and altitude 
with 
on side 
. Let 
be the midpoint of side , and let 
be the reflection of over . Let 
be a point on line 
such that lines 
and are parallel, and let the circumcircles of 
and 
meet again at 
. Prove that 
.
as the real axis. The vectors 
, 
and 
sum to 
, so the y-coordinates of 
sum to 0. Thus, the absolute value of one of them is the sum of the absolute values of the other two, hence done since all three triangles have as the base and some height away from the line.
lies on the opposite side of the Euler line, shown in blue, as 
and 
. (Note it is impossible for , , and to all be on the same side as the centroid is always inside the triangle.) We claim ![\[[AOH] = [BOH] + [COH].\]](http://latex.artofproblemsolving.com/8/f/3/8f314b79b8a65b59d482d4215c7a22e40a564189.png)
, , and onto the Euler line be 
, 
, and 
, respectively. The problem reduces to proving ![\[AA_1 = BB_1 + CC_1.\]](http://latex.artofproblemsolving.com/d/a/f/daf8edcd579fbd391b57259d5d44a826afd5cf9e.png)
as the midpoint of and 
as the projection of onto the Euler line. As 
is the midline in trapezoid 
, we have ![\[MM_1 = \frac{BB_1 + CC_1}{2}.\]](http://latex.artofproblemsolving.com/a/2/1/a218c71acc8d0586def08ede610e591b2c960f99.png)
and 
, where is the centroid and on the Euler line, give ![\[AA_1 = \frac{AG}{GM} \cdot MM_1 = 2MM_1 = BB_1 + CC_1. \text{ } \blacksquare\]](http://latex.artofproblemsolving.com/9/5/a/95aa012abd7a31e9d16efab571084c6041a15090.png)
![[asy]
size(250);
pair A, B, C, H, A1, B1, C1, M, G, M1;
A = dir(131);
B = dir(200);
C = dir(340);
H = orthocenter(A, B, C);
A1 = foot(A, H, 0);
B1 = foot(B, H, 0);
C1 = foot(C, H, 0);
M = .5B + .5C;
G = centroid(A, B, C);
M1 = foot(M, B1, C1);
markscalefactor=.005;
fill(A--A1--G--cycle, mediumgreen);
fill(G--M1--M--cycle, mediumgreen);
draw(rightanglemark(G, A1, A));
draw(rightanglemark(B, B1, A1));
draw(rightanglemark(G, M1, M));
draw(rightanglemark(M1, C1, C));
draw(A--B--C--cycle);
draw(B--B1);
draw(C--C1);
draw(M1--M--A--A1);
draw(B1--C1, blue);
label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$A_1$", A1, SE);
label("$B_1$", B1, NW);
label("$C_1$", C1, NE);
label("$M$", M, S);
label("$G$", G, NE);
label("$M_1$", M1, N);
dot(A);
dot(B);
dot(C);
dot(A1);
dot(B1);
dot(C1);
dot(M);
dot(G);
dot(M1);
[/asy]](http://latex.artofproblemsolving.com/6/5/6/65625d2e414ac13e8bc5f1399c4cabf940f0e922.png)
![$[AOH] = [BOH] + [COH]$](http://latex.artofproblemsolving.com/8/5/3/853f1072e0a7d60a5b52f1ef8a6ad6d342f36ca8.png)
. Now proceed with complex numbers. Then the area of ![$[AHO]$](http://latex.artofproblemsolving.com/8/0/0/800f77fab0baae6928d8a1124db82de0279bbc1b.png)
is given by,![\begin{align*}
\frac{i}{4}\begin{vmatrix}a&\frac{1}{a}&1\\ a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}&1\\ 0&0&1\end{vmatrix} = \frac{i}{4abc}\left [(a^2 - bc)(b+c)\right ]
\end{align*}](http://latex.artofproblemsolving.com/b/a/e/bae94743f54c57d7367176ab9026ab5d9550f917.png)
Similarly we can calculate the areas of ![$[BOH]$](http://latex.artofproblemsolving.com/3/d/6/3d69b0e59435695b9a852b712cc3e79d80ef144f.png)
and ![$[COH]$](http://latex.artofproblemsolving.com/9/4/c/94c96183afc1a0ff1b187093a79c3aed461ff843.png)
and adding gives the conclusion.
be on the opposite side of 
as , and . We claim that ![$[AOH]=[BOH]+[COH]$](http://latex.artofproblemsolving.com/5/4/6/5467395a75a1f5501213af3ccfcf4209033b3294.png)
Let be the centroid of 
Note that lies on since is the Euler line of . Let 
, 
, and 
be the projections of , , and respectively onto . It suffices to show 
. Let be the midpoint of and 
be its projection onto . Note that 
is the midline of trapezoid 
. By centroid 
ratios we have 
So , as desired.
, 
, and 
be the foots of the perpendiculars from , , and to , respectively. WLOG, let and lie on the same side of and lie on the other. Then, we want to prove that 
. Since the centroid lies on , this must be true by coordinates.
. lies on ., , and onto the Euler Line be , , and respectively. Then it is equivalent to show that 
. to be , and the projection of onto being 
.
. This follows easily from noticing that 
(by homothety) with a ratio, which finishes.
.![$$[AOH] = \frac12 (2R \cos A) (R) (\sin(|B-C|)) = \cos A \sin( B-C).$$](http://latex.artofproblemsolving.com/3/5/9/359202d64d5cb78c5d60d00352bd06199086a0bb.png)
Similarly, ![$$[BOH] = \cos B \sin(A-C), [COH] = \cos C \sin (A-B).$$](http://latex.artofproblemsolving.com/d/9/a/d9a15bdc0c31e8b94fe0cf5f70f6162689a50b9b.png)
But now simply from expanding, ![$[AOH] + [COH] = [BOH]$](http://latex.artofproblemsolving.com/0/2/d/02d37278f86705985e0a12b1d5b2eb750567db76.png)
. 
passes through the symmetric of one vertex in the midpoint of opposed side
is the farthest from it. It suffices to show that the distance from to the Euler line is equal to the sums of the distance from and to the Euler line. Let be the centroid of and let be the midpoint of 
.
, 
, 
, and 
as the projections of , , , and to the Euler line, respectively. Note that 
because 
with a ratio. Moreover, 
is the midline of trapezoid 
, so 
, as desired. 
![$$[AOH]=\frac {AO.AH.sin(B-C)}{2}=R^2cos(A).sin(B-C)$$](http://latex.artofproblemsolving.com/3/b/9/3b9d8c23fcd11e62319e4cf39ad5fb19cce433aa.png)
![$$[BOH]=\frac{BO.BH.sin(C-A)}{2}=R^2cos(B)sin(C-A)$$](http://latex.artofproblemsolving.com/f/d/1/fd1a33962b7a5973ed4d59b2cd748d3244ba5b2e.png)
![$$[COH]=\frac{CO.CH.sin(A-B)}{2}=R^2cos(C)sin(A-B)$$](http://latex.artofproblemsolving.com/0/c/0/0c0eba4b9ba474022574263bd964b090c9dc895e.png)
Then, ![$[AOH]+[BOH]+[COH]$](http://latex.artofproblemsolving.com/c/5/d/c5df177645a15d03b78801c4b1be55be7152588b.png)
=
to the Euler line, the largest one is the sum of the other two. But their average, is on the Euler line, so this is clear.
and lie on the same side of . Let 
be the foot to from 
respectively. We would like to show that 
. be the midpoint of and let be the foot from to 
. By the Thales' Theorem, 
is the Euler-line so it contains the centroid of . Now notice that 
.
, we obtain 
and we're done, as ![$[BOH] + [COH] = \frac{OH \cdot YB}{2} + \frac{OH \cdot ZC}{2} =
\frac{OH(YB + ZC)}{2} = \frac{OH \cdot AX}{2} = [AOH]$](http://latex.artofproblemsolving.com/3/f/0/3f0ee7192d1403c1854d21bf93c736a3fb36ebff.png)
. 
are roots of 
,
,
and 
is equal to the sum of the areas of the other two.
is the sum of the other two.
centered at 
.
of 
respectively. Moreover, it shrinks all distances by a factor of 
.
.
is half the distance 
,
are the images of 
respectively through 
.
,
.
Thus, 
a.s.o.![$16S\cdot [AOH]=(b^2+c^2-a^2)\cdot \left|b^2-c^2\right|$](http://latex.artofproblemsolving.com/a/d/f/adf178d68fa01c74575ad27618d046d3d3c67b16.png)
a.s.o. Prove easily that ![$a\le b\le c\Longrightarrow [BOH]=[AOH]+[COH]\ .$](http://latex.artofproblemsolving.com/9/f/4/9f4cb91360342ad33151e84e7f12eef345c1a5e2.png)
for angles ![$[OBH]=[OAH]+[OCH]$](http://latex.artofproblemsolving.com/9/a/9/9a91bd4c7c775446ceea0a8c93dfba1d412f8453.png)
.
in this case. Since 
and similarly for 
it is enough to show with 
area formula that![\[ \cos B \sin (A-C) = \cos A \sin (B-C) + \cos C \sin (A-B) \]](http://latex.artofproblemsolving.com/e/f/e/efe324d926f10a9a18b8d6a76d0f36bb207b3222.png)
![\[ \cos B (\sin A\cos C-\sin C \cos A)=\cos A(\sin B \cos C - \cos B \sin C)+\cos C(\sin A \cos B - \cos A \sin B) \]](http://latex.artofproblemsolving.com/a/0/1/a01d914d44c3e059f22caee36b6cf1d53e01928f.png)
which is true, so done.
,
,
,
.
,
and 
.![$$[AHO]=\frac{1}{2}
\left|
\begin{array}{ccc}
x_1 & y_1 & 1 \\
\sum x & \sum y & 1 \\
0 & 0 & 1
\end{array}
\right|=\frac{1}{2}(x_1\sum y-y_1 \sum x).$$](http://latex.artofproblemsolving.com/c/0/a/c0a41276752c06ed1fcc6bcbd1242fdd13b7c965.png)
Similarly, we can obtain that ![$[BHO]=\frac{1}{2}(x_2\sum y-y_2\sum x)$](http://latex.artofproblemsolving.com/3/c/3/3c3a074a774905cfa0eeb6a56c74d6afd36b7573.png)
and ![$[CHO]=\frac{1}{2}(x_3\sum y-y_3\sum x)$](http://latex.artofproblemsolving.com/e/1/1/e11e95965c872dd8fab86ca4098acc24e9ffa87a.png)
.![$$[AHO]+[BHO]+[CHO]=\frac{1}{2}(\sum x\sum y-\sum x\sum y)=0$$](http://latex.artofproblemsolving.com/d/8/d/d8d31cd3c56e05382d1fd3d50696f153e63b1713.png)
which immediately implies the desired conclusion.
,![\begin{align*}x\frac{[AXP]}{[ABC]} & =x\begin{vmatrix}1&0&0\\x&y&z\\p&q&r\end{vmatrix}=xyr-xzq, \\
y\frac{[BXP]}{[ABC]} & =y\begin{vmatrix}0&1&0\\x&y&z\\p&q&r\end{vmatrix}=yzp-yxr, \\
z\frac{[CXP]}{[ABC]} &=z\begin{vmatrix}0&0&1\\x&y&z\\p&q&r\end{vmatrix}=zxq-zyp,
\end{align*}](http://latex.artofproblemsolving.com/8/3/a/83ac693759fc41e5fd58cd6c48005e639fb35f2e.png)
so![$$x\frac{[AXP]}{[ABC]}+y\frac{[BXP]}{[ABC]}+z\frac{[CXP]}{[ABC]}=0,$$](http://latex.artofproblemsolving.com/a/8/3/a834840247504eb48cbda790ef0487638b83ebb2.png)
and therefore![$$x[AXP]+y[BXP]+z[CXP]=0,$$](http://latex.artofproblemsolving.com/3/0/c/30ca4fd252fda4a0c6b4cc869731d1bf2f4923ae.png)
as wanted.
,![\begin{align*}\sum \limits _{P\in \{A,B,C\}}[POH] & =\sum \limits _{p\in \{a,b,c\}}\frac{i}{4}\cdot \begin{vmatrix}p&\frac{1}{p}&1\\0&0&1\\a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}&1\end{vmatrix} \\
& =\sum \limits _{p\in \{a,b,c\}}-\frac{i}{4}\cdot \begin{vmatrix}p&\frac{1}{p}\\a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\end{vmatrix} \\
& =-\frac{i}{4}\cdot \sum \limits _{p\in \{a,b,c\}}\begin{vmatrix}p&\frac{1}{p}\\a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\end{vmatrix} \\
& =-\frac{i}{4}\cdot \begin{vmatrix}a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\\a+b+c&\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\end{vmatrix} \\
& =-\frac{i}{4}\cdot 0 \\
& =0.
\end{align*}](http://latex.artofproblemsolving.com/5/2/d/52d6df030a1a4cdd6bf54b95be42aa7dd451cd8a.png)
Done.
![$[AOH]+[BOH]+[COH]=3[GOH]=0$](http://latex.artofproblemsolving.com/c/8/5/c85fd5b3f9b445cb208bf28e80f1045761024858.png)
.