#18) Direct Pascal's Theorem
by Drunken_Master, Mar 7, 2018, 7:45 AM
Let 
and 
be the circumcircle and circumcenter of right triangle 
with 
. Let 
be any point on the tangent to at 
other than , and suppose ray 
intersects again at 
. Point 
lies on line 
such that 
. Prove that 
are collinear.
and 
be the circumcircle and circumcenter of right triangle 
with 
. Let 
be any point on the tangent to at 
other than , and suppose ray 
intersects again at 
. Point 
lies on line 
such that 
. Prove that 
are collinear.
.
,
and 
are collinear. It suffices to show that 
.
is a rectangle and 
,
This post has been edited 1 time. Last edited by pro_4_ever, Mar 8, 2018, 8:17 AM
be the circumcircle of the triangle 
and 
,
intersecting the interior of triangle 
.
,
,
,
,
and 
.
is midpoint of 
.
and 
are cyclic.
are collinear.
is 
s
,
Hence, 
:
Now, since 
is 
s
,
.
Hence, done!! 
,
,
,
be distinct circles such that 
,
,![\[
\frac{AB\cdot BC}{AD\cdot DC}=\frac{PB^2}{PD^2}.
\]](http://latex.artofproblemsolving.com/0/4/e/04ead9928d0c767a2209f610885e30f6b6e0a4e9.png)
denote image of 
under inversion.
is parallelogram, which implies 
,
is acute, with medians 
.
under reflection in 
;
is the image of 
.
and 
are concentric.
share the same perpendicular bisector, and so do 
.
and 
.
and 
be the points on 
and 
,
and 
.
and 
is on the circumference of triangle 
represent the circumcircle of triangle 
.
intersect 
.
and 
meet at 
.
.
is a kite with 
,
.
bisects 
shows that 
are collinear.
and 
.
.
,
,
,
, 
, 
.
to the circumcircle of 
s![\[b^2+c^2 = \tfrac12a^2 + 2AM^2 \iff AM = \tfrac12a = BM = CM\]](http://latex.artofproblemsolving.com/0/c/e/0ce867daae3e1eac85e7e2d65e9be1cdfc1e5b2c.png)
be the line passing through the midpoint of segment 
.
lies on 
and 
.
and 
with 
be the radius of 
which gives 
.
.![\[2=\frac{YD}{YX}=\frac{DO}{OM}=\frac{R}{OM}.\]](http://latex.artofproblemsolving.com/1/1/c/11ceae2c410f6eaa11fa63db9b7c7ebf1276fd0f.png)
So, 
.
meet the circle again at 
is a rhombus. So, ![\[180^{\circ}-\frac{\angle BOC}{2}=180^{\circ}-\angle BDC=\angle BAC = \angle BA'C = \angle BOC.\]](http://latex.artofproblemsolving.com/7/b/0/7b0645de1dc11320e6ac37def30e5bdefd8d8307.png)
So, 
.
,
be the point on the segment 
such that 
.
.
and 
.
.
, and as 
. Let 
. We get the displacement vector 
. The condition that length of 
. Now observe that only the positive root satisfies 
,
, where we defined 
.
, and analgously, 
, hence 
. Now, the displacement vector (after normalization) equates to 
. Now, we need to show that the lenght of this displacemnt vector is symmetric in 
and 
,![$$ Length[MN]^2 = \frac{b^2d^2}{(c^2+d(a+b)+x^2)^2}[\frac{Terms}{ (c^2+x^2+ad)^2 } ]$$](http://latex.artofproblemsolving.com/9/9/3/993362f4272a430655e3c7698a5b8b6eae7ef7c0.png)
, where 
![$$ = a^2 [ (a^2+b^2+x^2+ad)^2 - b^4-b^2a^2-2b^2x^3+b^2d^2] = a^2[(c^2+x^2+ad)^2 - b^2(b^2+a^2+2x^2-d^2)] = a^2[(c^2+x^2+ad)^2 - b^2(d^2-d^2)] = a^2(c^2+x^2+ad)^2$$](http://latex.artofproblemsolving.com/b/5/3/b5321d5e1e51f787c8c6bf10745668b9680e2b07.png)
, so we finally get ![$$ Length [MN]^2 = \frac{b^2d^2}{(c^2+d(a+b)+x^2)^2}[\frac{Terms}{ (c^2+x^2+ad)^2 } ] = \frac{b^2d^2}{(c^2+d(a+b)+x^2)^2}[\frac{a^2(c^2+x^2+ad)^2}{ (c^2+x^2+ad)^2 } ] = \frac{a^2b^2d^2}{(c^2+d(a+b)+x^2)^2}$$](http://latex.artofproblemsolving.com/d/6/6/d66a4469629d93af5941d5b1eec6585529fe73ff.png)
, and as of course it's symmetric in 
and the circumcenter 
prove that this triangle is unique and one of it's angles is 
![[asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(14cm); 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 = -11.56, xmax = 11.56, ymin = -8.04, ymax = 8.04; /* image dimensions */pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); /* draw figures */draw(circle((-1.47,0.54), 6.618194617869741), linewidth(0.8) + linetype("4 4") + rvwvcq); draw((3.3824279616279336,5.040493625949423)--(-7.16,-2.84), linewidth(2) + wrwrwr); draw((-7.16,-2.84)--(4.307449273562116,-2.688247185612742), linewidth(2) + wrwrwr); draw((3.3824279616279336,5.040493625949423)--(4.307449273562116,-2.688247185612742), linewidth(2) + wrwrwr); draw(circle((1.2264297598997855,0.009779353729876573), 2.7385589808979787), linewidth(1.2) + dtsfsf); draw((xmin, 2.333357360032656*xmin-2.8519195530953714)--(xmax, 2.333357360032656*xmax-2.8519195530953714), linewidth(1.2) + linetype("4 4") + sexdts); /* line */ /* dots and labels */dot((-1.47,0.54),dotstyle); label("$O$", (-1.38,0.74), NE * labelscalefactor); dot((-7.16,-2.84),dotstyle); label("$B$", (-7.08,-2.64), NE * labelscalefactor); dot((4.307449273562116,-2.688247185612742),dotstyle); label("$A$", (4.38,-2.48), NE * labelscalefactor); dot((3.3824279616279336,5.040493625949423),dotstyle); label("$C$", (3.46,5.24), NE * labelscalefactor); dot((1.2264297598997855,0.009779353729876573),linewidth(4pt) + dotstyle); label("$I$", (1.3,0.16), NE * labelscalefactor); dot((3.469877235190048,-1.5677535596633183),linewidth(4pt) + dotstyle); label("$H$", (3.54,-1.4), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]](http://latex.artofproblemsolving.com/8/9/1/891b4de8fa68ae5581ff80ecd7f93e77598400c9.png)
.
Let 
be the reflection of 
Note that 
since 
Also, 
which means 
Therefore, 
and 
)
be the relfection of 
Since in this case 
because 
by assumption of this case. Which is absurd because 
intersects 
Hence this case is not possible.
so they have a side of equal length. Place the equal sides as chords of the circle such that they coincide, let the chord be 
![\[\angle AIB=\angle 90^{\circ}+\frac{60^{\circ}}{2}=120^{\circ}=\angle AOB\]](http://latex.artofproblemsolving.com/0/b/0/0b0cf8b2055883b2c2c664182c0d684296f9c456.png)
Hence 
So as we move 
of 
Drop a perpendicular from 
(![\[R^2-2Rr=r^2\implies r=\mbox{ fixed quantity in terms of R} (1)\]](http://latex.artofproblemsolving.com/4/b/8/4b8a075dab2bf8bb03160a83d0c6207e19158584.png)
Since we fixed the circumradius of both the triangles, hence 
of 
twice(it is more easier if we drop a perp from 
be feet of perpendiculars from 
and 
be diametrically opposite points of 
and 
respectively. Prove that 
where 
Similarly,![\[\frac{CF}{FA}=\frac{CD\cos C}{b-CD\cos C}\]](http://latex.artofproblemsolving.com/f/f/d/ffd59570fcac16d431a85c6413034a124653d68e.png)
Applying Ceva's theorem,
We also have ![\[CD+BD=a.\]](http://latex.artofproblemsolving.com/c/e/6/ce65ca77138d77458a58e29765370fd298252086.png)
Therefore,![\[BD=\dfrac{a\cos B\cos C-b\cos B+c\cos C}{2\cos B\cos C}\qquad (\spadesuit).\]](http://latex.artofproblemsolving.com/a/a/e/aae508a18636cbd55880feadcb439858d5040fee.png)
it follows that![\[\angle ABY=90^{\circ}-\angle B,\qquad \angle BAY=\angle EDA=90^{\circ}-\angle BAD\]](http://latex.artofproblemsolving.com/2/f/1/2f121dfa27a591ba29cc46bcb54ee8a86ea1c51f.png)
Using Sine Law in ![\[\frac{c}{\sin\angle BYA}=\frac{YB}{\cos\angle BAD}\]](http://latex.artofproblemsolving.com/8/1/1/811abf1229b8e89caa7de668c7f8662477aa7081.png)
Therefore,
Now, note that,![\[\cos B=\frac{c^2+BD^2-AD^2}{2c\cdot BD}\implies AD^2-BD^2=c^2-2c\cdot BD\cos B\]](http://latex.artofproblemsolving.com/8/6/3/86353bd7471d4cf8a2af338a78e62a25d302fda9.png)
Substituting the above in 
Consider,
W.L.O.G. assume 
Then 
Putting the above in the denominator of 
Thus, ![\[YB=\frac{\Delta}{b\cos C}.\]](http://latex.artofproblemsolving.com/6/b/5/6b5701a3a03ef421221c15e6935e2165e57e1c77.png)
Let 
Then as 
,![\[PB=\frac{a\cdot\tfrac{bc}{4R}}{CA'}=\frac{\Delta}{CA'}=\frac{\Delta}{b\cos C}\]](http://latex.artofproblemsolving.com/0/f/1/0f12c7804ab3b5dadfca6dfc5b6eec66ce13f60d.png)
Thus we obtain 
Hence 
And we are done. ![[asy]import olympiad;
unitsize(1.5cm);
pair A=(0.8,2), B=(0,0), C=(3, 0), D=relpoint(B--C, 0.8), F = foot(D, A, B), E = foot(D, A, C), O_1 = circumcenter(A,B,D), O_2 = circumcenter(A,D,C), Y = rotate(180,O_1)*D, X=rotate(180,O_2)*D, Z, K = foot(A,B,C);
draw(A--B--C--cycle);
draw(D--E, dashed+0.8blue); draw(D--F,dashed+0.8blue);
draw(rightanglemark(D,E,C,4),0.8blue); draw(rightanglemark(D,F,B,4),0.8blue);
path U = circumcircle(A,B,D);
path V = circumcircle(A,C,D);
draw(U, linetype("3 4")+red);draw(V,linetype("3 4")+red);
draw(B--E, linetype("2 4 6 4")+0.7green); draw(C--F, linetype("2 4 6 4")+0.7green); draw(A--D, linetype("2 4 6 4")+0.7green);
dot(intersectionpoint(B--E, C--F),0.7green);
Z = intersectionpoint(B--X,C--Y);
draw(B--X, orange);draw(C--Y, orange);draw(A--K, orange);
draw(rightanglemark(A,K,C, 4), orange);
dot(A);dot(B);dot(C);dot(D,0.8red);dot(E,0.8blue);dot(F,0.8blue);dot(X,0.8red);dot(Y,0.8red);dot(Z, orange);dot(K,orange);
label("$A$", A, dir(120));
label("$B$", B, SW);
label("$C$", C, SE);
label("$D$", D, dir(-75), 0.8red);
label("$E$", E, dir(45),0.8blue);
label("$F$", F, dir(135),0.8blue);
label("$X$", X, dir(45), 0.8red);
label("$Y$", Y, dir(135), 0.8red);
label("$A'$", K, S, orange);
label("$M$", Z, dir(20), orange);[/asy]](http://latex.artofproblemsolving.com/5/b/5/5b5c4976812577ada4205dacc53e072392d87212.png)
March 2018
February 2018