fair coins are flipped. If 
of them land heads, find the expected value of 
.
and 
.
Let 
.
Prove that
Y by Davi-8191, megarnie, Quidditch, HWenslawski, mathlearner2357, Adventure10, Mango247
Let 
be an acute triangle, with 
as its circumcenter. Point 
is the foot of the perpendicular from 
to line 
, and points 
and 
are the feet of the perpendiculars from to the lines 
and 
, respectively.
Given that
prove that the points 
and are collinear.
be an acute triangle, with 
as its circumcenter. Point 
is the foot of the perpendicular from 
to line 
, and points 
and 
are the feet of the perpendiculars from to the lines 
and 
, respectively.Given that
prove that the points 
and are collinear.
cyclic we have ![\[\angle ABH=\angle AHP=\angle AQP.\]](http://latex.artofproblemsolving.com/5/5/9/559aa86fc42a0797f25eab4aaea02c221f6fb75d.png)
Now I shall show that 
,
.
is true since 
,
and we can finish easily.![\[\dfrac{AQ}{AB}=\dfrac{AQ}{AH}\cdot\dfrac{AH}{AB}=\sin B\sin C.\]](http://latex.artofproblemsolving.com/a/d/5/ad5c2219a3e38f284e8ac23a58eb0f13fb8abf67.png)
Now note that by expressing the area of ![\[\dfrac12a\cdot (AH)=\dfrac12a\cdot (R\sqrt 2)=\dfrac{abc}{4R}\quad\implies\quad 2R^2\sqrt 2 = bc.\]](http://latex.artofproblemsolving.com/3/4/6/34693efc218b234626864a53c5cf87134b439a08.png)
But note that by Law of Sines 
Setting these equal yields 
,
(just use similar triangles or 
inversion). Then by Power of a Point,
Consider the transformation 
which dilates 
from 
and reflects about the 
clearly lies on 
,
so 
,
are collinear, as desired.
,
,
,
.
,
are tangent at 
where 
is the 
,
,
,
so 
are cyclic but note that 
so 
,
![$[AOP] + [AOQ ] = [APQ]$](http://latex.artofproblemsolving.com/f/8/4/f84c2f3b1d2d40de625ebbd3b0e7a89c287c732b.png)
,
as 
.
is obviously equivalent to this as well, so we are done.
![$[AOP] + [AOQ ] = [APQ],$](http://latex.artofproblemsolving.com/3/c/a/3cadbc66a1946d6d053fb4e799a0d20dd75647e7.png)
and using the sine area formula gives 
Since 
and similarly 
Since 
we simplify our equation to 
In right triangle 
observe that 
so 
but right triangle 
gives 
hence equating gives 
by the condition. Symmetrically, 
Substituting back in, 
which directly simplifies to 
where the last step follows by the Extended Law of Sines. Now, expressing the cosines of the angles in terms of side lengths using the Law of Cosines and substituting, it remains to show 
which is trivial upon simplification.
by similar triangles. Let the inverse of 
.
so 
the antipode of 
which means that 
be the circumradius, the given length condition is 
.
and 
so that 
and 
.
and similarly 
.
are collinear, it suffices to show that ![$[APQ] = [AOP] + [AOQ]$](http://latex.artofproblemsolving.com/5/3/7/5374617b6e5f0cc1493ecf539d2a486c35aaa20c.png)
.![$[APQ]$](http://latex.artofproblemsolving.com/8/e/9/8e93886076372bdc243166123cdf6461b19d7916.png)
.
and 
,![\[ [APQ] = \frac{1}{2}(\sqrt{2}R\cos x)(\sqrt{2}R\cos y)\sin (x + y) = R^2\cos x\cos y\sin (x + y).\]](http://latex.artofproblemsolving.com/9/1/a/91a285ce5a4ffbd04b309b07ee5b66b3579392d3.png)
Now, noting that 
,![\[ [AOP] = \frac{1}{2}AP\cdot AO\sin \angle OAP = \frac{\sqrt{2}}{2}R^2\cos x\sin y,\]](http://latex.artofproblemsolving.com/0/b/4/0b4c0f5f1e7209bfdafb070c9e50bfee216e9e83.png)
and similarly ![$[AOQ] = \tfrac{\sqrt{2}}{2}R^2\cos y\sin x$](http://latex.artofproblemsolving.com/f/a/c/fac850d373b71abfb01e129c97b747acd1b58e51.png)
.![$[AOP] + [AOQ] = \frac{\sqrt{2}}{2}R^2\sin (x + y)$](http://latex.artofproblemsolving.com/f/4/c/f4ca479df712b1b89d7d18d96def708488afb8b9.png)
,
.![\[ [ABC] = \frac{1}{2}AH\cdot BC= 2R^2\sin A\sin B\sin C,\]](http://latex.artofproblemsolving.com/4/c/9/4c9c9f48c1ff21578e7dd98155ee332a0716299a.png)
![\[ \frac{1}{2}\sqrt{2}R\cdot 2R\sin A = 2R^2\sin A\sin B\sin C\implies \sin B\sin C = \frac{\sqrt{2}}{2}.\]](http://latex.artofproblemsolving.com/b/a/6/ba6a4a667abed0508e3ba60c6f2d55a956791af2.png)
However, 
,
and we are done.
be the intersection of line 
and 
because 
.
because of circumcircle properties. Thus, 
.
is the altitude.![\[\frac{AQ}{AB} = \frac{AO_{1}}{AH}\]](http://latex.artofproblemsolving.com/a/2/7/a2742d4f5122f82d674d02af07a51f6fc1016e2c.png)
![\[AQ = AH\sin{C}\]](http://latex.artofproblemsolving.com/a/f/1/af1bfeb97d98d07d0991ad15bc5d31d8ca6c4802.png)
and![\[\frac{AB}{\sin{C}} = 2AO\]](http://latex.artofproblemsolving.com/5/2/9/52911a8825b2d34bf6fb6f71cf82554d7801692e.png)
Thus,
Thus, 
proving that 
which readily implies the cyclic quads 
as desired. 
.
in a similar fashion so we are done. 
![[asy]
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[/asy]](http://latex.artofproblemsolving.com/b/e/5/be537f65e4fe9879d7f0968bd01d59a958527b2a.png)
,
lying on the unit circle.![\[h=\frac{1}{2}\left(a+b+c-\frac{bc}{a}\right).\]](http://latex.artofproblemsolving.com/2/9/e/29e9932174dc13e9118604e0e43c83f20887735e.png)
Hence, the given condition reduces to
Then, by complex foot again, we have![\[p=\frac{1}{2}\left(a+b+h-ab\overline{h}\right),\]](http://latex.artofproblemsolving.com/f/4/f/f4f2f1d7c8e45dac429675cdc936784fe9290348.png)
which expands to
Analogously, we have![\[q=\frac{1}{4}\left(2a+2c+b-\frac{bc}{a}-\frac{ac}{b}+\frac{a^2}{b}\right).\]](http://latex.artofproblemsolving.com/2/5/4/254f90705f8363a906dbe6b9b95dffde6c071c22.png)
By the complex collinearity criterion, it suffices to show that![\[\frac{0-p}{0-q}=\frac{p}{q} \in \mathbb{R}.\]](http://latex.artofproblemsolving.com/7/e/8/7e87d25a5ef4cf191a885cb8e51a67a0f00d1cc1.png)
Thus, we calculate
Now, let 
and 
be the numerator and denominator of this fraction, respectively, so that 
.![\[x+\overline{x}=4+\left(-\frac{a^2}{bc}-\frac{bc}{a^2}+\frac{2a}{b}+\frac{2b}{a}+\frac{2a}{c}+\frac{2c}{a}-\frac{b}{c}-\frac{c}{b}\right)=4-4=0\]](http://latex.artofproblemsolving.com/6/5/1/65165df9819776840f5746ef7d4b3674a4e61e28.png)
by substituting directly into the result we derived from the given condition. Similarly, we get 
.![\[\frac{x}{y}=\frac{-\overline{x}}{-\overline{y}}=\frac{\overline{x}}{\overline{y}} \implies \frac{x}{y} \in \mathbb{R},\]](http://latex.artofproblemsolving.com/1/6/d/16d703fd3b0f009a32827cc317279f2da22e591d.png)
which clearly finishes.
,
(inversely similar.) Now
Therefore, 
,
with respect to 
which yields 
as desired.
and 
.
we can use:
to a line 
be 
.
Thus we shall get, 
and 
.
Now clearly we have 
and 
.
Now we need to find the equation of line 
and 
be the slope of line 
Now we can find the equation of line 
Now we can just plug the coordinates of 
This condition is exactly equivalent to that of 
and hence, we are done. 
or 
by a cyclic quadrilateral. Because 
or eqiuvalently 
ergo 
So from our condition we just need to prove 
and we are done.
and 
by basic right triangles. Plugging this into our given condition we get 
which reduces to 
, done.
.
Note that 
Therefore by right triangle ratios 
.
