,
,
,
,
or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.
,
isn't even rigorously defined in this expression. What exactly do we mean by 
.
,
does not exist in the first place, although you will see why in a calculus course. So the point is that 
?
indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that 
,
or ![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function 
given by the mapping 
is undefined for 
.
is undefined because dividing by ![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In ![\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by ![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by ![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and 
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with ![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by 
be its altitudes. Let 
be its orthocenter, and let 
and 
be the circumcenters of triangles 
and 
.
be the second intersection of the circumcircles of triangles 
and 
.
,
,
-
are concurrent.
Prove that: 
is the sum of 
square numbers
Let 
and 
is the sum of 
is the sum of 
and circumcircle 
.
.
and 
,
is tangent to 
,
.
and 
on 
and 
are on the same side of 
be midpoint of segment 
be midpoint of segment 
with 
and 
and 
are numbers that work, but 
and 
don't work.)
Y by buratinogigle, rightways, YadisBeles, Kezer, Davi-8191, R8450932, UpvoteFarm, Adventure10, Mango247, Rounak_iitr
We say that a triangle 
is great if the following holds: for any point 
on the side 
, if 
and 
are the feet of the perpendiculars from to the lines 
and 
, respectively, then the reflection of in the line 
lies on the circumcircle of the triangle . Prove that triangle is great if and only if 
and 
.
Senior Problems Committee of the Australian Mathematical Olympiad Committee
is great if the following holds: for any point 
on the side 
, if 
and 
are the feet of the perpendiculars from to the lines 
and 
, respectively, then the reflection of in the line 
lies on the circumcircle of the triangle . Prove that triangle is great if and only if 
and 
.Senior Problems Committee of the Australian Mathematical Olympiad Committee
This post has been edited 1 time. Last edited by MellowMelon, May 18, 2017, 3:29 AM
Reason: add proposer
Reason: add proposer
the reflection of 
.
and 
,
.
is a convex quadrilateral. Since we want this point to also lie on the circumcircle of 
.
,
,
.
.
.![\[\angle BD'A = 180^\circ-\angle ACB=180^\circ-\angle ADQ=180^\circ-\angle APQ=180^\circ-\angle D'AP,\]](http://latex.artofproblemsolving.com/1/0/e/10e3da317a40807749285cf509eacf4ae329a189.png)
which contradicts the fact that 
is non-degenerate.
or 
.
is a square, so 
is also the angle bisector from 
as desired.
- this is because 
.
since 
- hence 
as well. This implies that 
.
.![\[\angle BD'D=\dfrac{\angle BPD}2 = \dfrac{90^\circ}2 = 45^\circ\qquad\text{and}\qquad \angle CD'D = \dfrac{\angle CQD}2 = \dfrac{90^\circ}2 = 45^\circ.\]](http://latex.artofproblemsolving.com/f/c/9/fc98d3fafd804a70a4eb0bbcebeaccca5ce6fede.png)
Hence 
and indeed 
.
by angle chasing. Now use the first observation and choose 
and 
Let 
denotes the foot from 
and 
denote the reflection of 
it's easy to see that 
is the orthocenter of 
which implies that 
are symmetry over the perpendicular bisector of ![\[D^*\text{ lies on the circumcircle of }\triangle ABC\iff\angle BD^*C=\angle BE^*C=\angle A\iff E^*\equiv A.\]](http://latex.artofproblemsolving.com/c/3/6/c36b6ccdf59ee9ff74c0459f3264dcdb7f216206.png)
Therefore 
This is because 
and 
![[asy]
size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt);
void b(){
pair A=D("A",dir(110),dir(110));
pair B=D("B",dir(-150),dir(-150));
pair C=D("C",dir(-30),dir(-30));
pair D=D("D",extension(A,origin,B,C),S);
pair P=D("P",foot(D,A,B),W);
pair Q=D("Q",foot(D,A,C),E);
pair E=D("E",foot(A,B,C),S);
pair D1=D("D^*",2*foot(D,P,Q)-D,dir(120));
pair E1=D("E^*",2*foot(E,P,Q)-E,N);
D(A--B--C--cycle);
D(P--D--Q,magenta);
D(B--E1--C);
D(B--D1--C);
D(P--Q,dashed);
draw(anglemark(B,E1,C,3),deepgreen);
draw(anglemark(B,D1,C,3),deepgreen);
D(E1--E,dotted);
D(D1--D,dotted);
D(circumcircle(A,P,Q),red+linetype("4 4"));
}
b(); pathflag=false; b();
[/asy]](http://latex.artofproblemsolving.com/5/e/5/5e56a5b11abd0856f6c7de39c325411fa498a000.png)
and![\[\begin{aligned}D^* \text{ lies on the circumcircle of }\triangle ABC &\iff D^* \text{ sends $\triangle D^*PB$ to $\triangle D^*QC$}\\
&\iff \frac{DP}{PB}=\frac{D^*P}{PB}=\frac{D^*Q}{QC}=\frac{DQ}{QC}\\
&\iff \tan{\angle B}=\tan{\angle C}\\
&\iff \angle B=\angle C. \end{aligned}\]](http://latex.artofproblemsolving.com/7/2/c/72c9918701003856b554bde67c5db31bb1d94b67.png)
as desired. 
![[asy]
size(7cm); pointpen=black; pathpen=black; pointfontpen=fontsize(9pt);
void b(){
pair A=D("A",dir(70),dir(70));
pair B=D("B",dir(180),W);
pair C=D("C",dir(0),E);
pair D=D("D",foot(A,B,C),S);
pair P=D("P",foot(D,A,B),W);
pair Q=D("Q",foot(D,A,C),S);
pair D1=D("D^*",2*foot(D,P,Q)-D,dir(70));
D(P--D--Q,magenta);
D(D--D1,dotted);
D(B--P--D1--cycle,blue+dashed);
D(D1--Q--C--cycle,blue+dashed);
D(A--P--Q--cycle);
D(B--C);
}
b(); pathflag=false; b();
[/asy]](http://latex.artofproblemsolving.com/9/7/a/97aa351d7f6107d1b9e8e7cc7d748e99f92506fa.png)
by setting 
.
are the midpoints of 
respectively as 
is cyclic, thus 
.
is tangent to 
at 
,
and they are on same side of 
,
,
.
.
,
.
be projections of 
on 
is isosceles trapezoid, so 
so 
.
and 
is cyclic. 
,
.
such that ![\[f(x)+f(y)=x^2+y^2\]](http://latex.artofproblemsolving.com/a/f/9/af9cf6c9505676db971e5eec3cadadee118cb09d.png)
for all real 
and 
.
to obtain ![\[2f(x) = 2x^2\quad\implies\quad f(x) = x^2.\]](http://latex.artofproblemsolving.com/4/9/d/49d92ef58916aba960914e76165520920917af5e.png)
Now we just need to check that 
always works, which it does.
is the only possible function which satisfies the given conditions, all we need to do is to look at the case where 
.
is cyclic, so 
.
and 
gives 
,
.
is cyclic. We are done.
,
cyclic gives 
,
is a square, so 
.
is equivalent to 
,
.
as 
,
.
is cyclic, so this is equivalent to the condition![\[ \frac{D'P}{PB} = \frac{DP}{PB} = \frac{DQ}{QC} = \frac{D'Q}{QC}\]](http://latex.artofproblemsolving.com/a/0/8/a08898590da697cb79b19113e2abacca2c33563b.png)
However, we have![\[ \frac{DP}{PB} = \frac{QC}{DQ}\]](http://latex.artofproblemsolving.com/4/d/a/4da9743884d331716b747b28437cf0809f211b8d.png)
since 
,
and 
being isosceles right. But they are similar to the original, so the original condition is equivalent with 
and 
simplifies the problem quite well, and it is easier to interpret reflection in tje Cartesian plane. I will post the solution as soon as I find one.
Angle Bisector of 
lies on 
lie on opposite sides if 
lie on same side of 
which means that
.
.
.
are circumcenter of 
and 
respectively.
,
note 
,
.
.
,
and 
.
,
is cyclic, so 
,
,
is congruent to 
,
.
.
,
,
,
are the midlines of 
.
,
and 
,
.
,
let 
and 
be the feet from 
lies on 
Hence, 
and since 
is cyclic, 
and 
Let 
be the foot from 
lies on 
and 
Hence, 
is a square and 
so 
Similarly, 
so 
,
,
which means 
so 
is cyclic with diameter 
implies 
is a cyclic isosceles trapezoid. Hence, 
and 
.
which means the only possible solution is 
and 
are also isosceles right triangles, we have 
so 
Thus, 
follows from 
,
and 
respectively, and the reflection 
or 
,
.
.
,
,
and when 
,
.
,
.
are equidistant from 
is a diameter of 
so it follows that 
.
,
are isogonal in 
.
,
.
,
lie on the same side of 
,
as well. Thus 
,
.
,
is cyclic, so 
implies 
,
implies 
,
implies 
bisects 
