Let $ABC$ be the triangle. We start by constructing triangles congruent to $ABC$ on its sides as in the figure below.

We can see that $\angle YAZ =$$ \angle YAC + \angle CAB + \angle BAZ =$$ \angle BCA + \angle CAB + \angle ABC =$$ 180^{\circ}$ from the congruent triangles. This implies that $Y,A$ and $Z$ all lie on a line. The fact that $ \angle ABC = \angle BAZ$ gives that $YZ$ is parallel to $BC$ due to alternate angles. This, together with the fact that segments $AY$ and $AZ$ are equal because of the congruent triangles, means that the altitude from $A$ to $BC$ is also the perpendicular segment bisector of $YZ$.

We can of course do the same calculations for the other two altitudes of triangle $ABC$ and find that they also are perpendicular bisectors of the sides of triangle $XYZ$. Because we know that these perpendicular bisectors all meet at the center of the circumscribed circle of triangle $XYZ$, the altitudes of triangle $ABC$ are concurrent.

Simply notice that the lines $AH, AB$ and $AC$ are all altitudes of the triangle $HBC$ because of the right angles. Since $A$ is the intersection of these three altitudes, it must be the orthocenter of triangle $HBC$.

When the triangle is acute, all the altitudes meet the opposite segments and the orthocenter lies inside the triangle. 

On the other hand, if the triangle is obtuse, then two of the altitudes never pass through the inside of the triangle. Therefore the orthocenter must lie outside the triangle if it is obtuse.

The orthocenter lies on the triangle itself if and only if it is a right triangle. In this case two of the altitudes of the triangle meet at the vertex opposite to the hypotenuse.

First consider the case when $ABC$ is acute.

Because $\angle AFH = \angle AEH = 90^{\circ}$, two opposite sides in quadrilateral $AFHE$ add up to $180^{\circ}$ and so $AFHE$ is inscribed in a circle. $BCEF$ is also inscribed in a circle because of the inverse of the inscribed angle theorem, as $\angle BFC = \angle BEC = 90^{\circ}$. Using similar arguments it can be shown that the $4$ other quadrilaterals

$$BDHF, CAFD, CEHD \text{ and } ABDE$$

are also inscribed in circles. If the triangle is obtuse we instead get the following situation.

Note that for quadrilateral $ABDE$ we here have to use the sum of opposite angles instead of the inverse of the inscribed angle theorem to show that it is inscribed in a circle. Similar adjustments also have to be done for some of the other quadrilaterals, but the circles are still there.

We know that $BCEF, CEHD$ and $BDHF$ are inscribed quadrilaterals.

If $\angle HDE = x$, we then have that $\angle HCE = x$ from inscribed angles in $CEHD$. This means that $\angle FCE = x$, implying that $\angle FBE = x$ because of inscribed angles in $BCEF$. From this, we see that $\angle FBH = x$ which means that $\angle FDH = x$ from inscribed angles in $BDHF$.

On the whole, we then have that $DH$ is the angle bisector of $\angle FDE$. By similar calculations we can see that $EH$ and $FH$ are also angle bisectors in triangle $DEF$. Since the angle bisectors meet at the incenter, $H$ must be the incenter of triangle $DEF$.

Consider the intersection $G'$ of the medians $BM$ and $CN$.

We first notice that $|AN|=|NB|$ and $|AM|=|MC|$ from the definition of a median. This implies that $\frac{AB}{AN}=2=\frac{AC}{AM}$, which gives that triangles $ABC$ and $ANM$ are similar by side-angle-side similarity. Furthermore the ratio of the similarity is $2:1$. Because of the similar triangles we can see that $\angle AMN = \angle ACB$, which means that $BC$ and $MN$ are parallel. We also know that $|BC|=2|MN|$ from the ratio of similarity. Alternate angles give that 

 $$\angle BCN = \angle CNM \iff \angle BCG' = \angle MNG'.$$

This together with $\angle MG'N = \angle GG'C$ from vertical angles implies that triangles $BCG'$ and $MNG'$ are similar from angle-angle similarity. The ratio of similarity is again $2:1$ since $|BC|=2|MN|$, and we thus have that 

 $$\frac{BG'}{G'M} = 2 = \frac{CG'}{G'N}.$$

By similar calculations, all pairs of medians intersect each other in points that divide them in the ratio $2:1$. But because every median just has one such point, it must intersect the other two medians at the same point. This implies that all three medians are concurrent. It is then also clear that this point divides the medians into segments of the ratio $2:1$ and our proof is complete.

Consider the triangle $ABC$ with medians $AX, BY$ and $CZ$, and centroid $G$.

There are many ways of approaching this problem, but one of them is to use the ratio of the medians that we just established. We can see that triangles $AGY$ and $CGY$ have the same area since they share a common base ($|AY|=|CY|$) and altitude (from $G$). Call this common area $T$ and notice that triangle $AGC$ then has area $2T$. Triangle $XGC$ shares an altitude from $C$ with $AGC$, but has half the base because $AG:GX=2:1$. This means that triangle $XGC$ has half the area of $AGC$, that is $\frac{1}{2} \cdot 2 T=T$. Similarly we can show that triangles $AGZ, BGZ$ and $XGB$ all have equal area, and the conclusion follows from that fact that triangles $XGC$ and $XGB$ have the same area. This can be seen by the fact that $|XC|=|XB|$ and that they have a common altitude from $G$.

The correct answer is $\textbf{(C)}$. We first try to find out what kind of curve the centroid traces out. Call the center of the circle $O$.

It is clear that $O$ is the midpoint of the diameter $AB$ and so the radius $OC$ is a median of the triangle $ABC$. If we call the centroid $G$, we also know that $G$ lies on $OC$ and that $|CG|=2 |OG|$. The ratio implies that 

$$2|OG| + |OG| = |CG| + |OG| \iff 3|OG| = |OC| \iff |OG| = \frac{|OC|}{3}.$$

Now, since $OC$ is a radius of the circle we can see that $|OC|=12$ and we thus have $|OG|= \frac{12}{3} = 4$. This means that as $C$ moves around the circle, $G$ traces out the points at a constant distance of $4$ from $O$. This shape is of course a circle with radius $4$ and we are almost done. Simply notice that the area of this circle is $\pi \cdot 4^2 \approx 50$ which means that $\textbf{(C)}$ is the correct answer.

It turns out that the centroid of triangle $XYZ$ is simply the centroid of triangle $ABC$. This can be guessed by inspection of the figure below and there are multiple ways to show it.

One way is to notice that $XY$ is parallel to $AB$ and $ZX$ is parallel to $CA$ (we established this in our proof of the existence of the centroid). This means that $AZXY$ is a parallelogram with diagonals $AX$ and $YZ$. Since diagonals in a parallelogram bisect each other, we then know that $AX$ bisects the segment $YZ$. This implies that $AX$ is a median in triangle $XYZ$, and in a similar way, we find that $CZ$ and $BY$ are also medians in triangle $XYZ$. Since they are medians in both of the triangles $ABC$ and $XYZ$, their intersection is the centroid for both of these triangles as desired.

We note that $\angle SBC = \angle SAC$ and $\angle BCS = \angle BAS$ from the inscribed angle theorem. Since $\angle SAC = \angle BAS$ from the definition of an angle bisector, we get that $\angle SBC = \angle BCS$, implying that $BCS$ is isosceles.

For simplicity we use $\angle A$ to denote $\angle BAC$, and we use $\angle B$ and $\angle C$ in a similar way.

Angle chasing gives that 

$$\angle SBI = \angle SBC + \angle CBI = \angle SAC + \angle CBI$$

from the inscribed quadrilateral $ABSC$, and since 

$$ \angle SAC + \angle CBI = \frac{1}{2} \angle A + \frac{1}{2} \angle B$$

from the angle bisectors, we have that $\angle SBI = \frac{\angle A + \angle B}{2}$. But we also have that 

$$\angle BIS = 180^{\circ} - \angle BIA = \frac{\angle A + \angle B}{2}$$

by the angle sum in triangle $ABI$. This means that $\angle SBI = \angle BIS$ and so the triangle $SBI$ is isosceles.

Make sure that you have drawn your triangle in a circle. Then simply construct one angle bisector and draw it all the way until it intersects the circle again. Take your compass, place it at this intersection point and adjust the other tip so that it lies on either of the two closest vertices of the triangle. Then draw this circle, or a relevant portion of it, and mark the intersection of the circle with the angle bisector (the intersection inside the triangle). The superlemma ensures us that this is the incenter.

The superlemma gives that the centers of the circumscribed circles of $\triangle IAB, \triangle IBC$ and $\triangle ICA$ are just the intersections of the angle bisectors with the circumscribed circle of $ \triangle ABC$. This means that the centers lie on the circumscribed circle of $\triangle ABC$, whose center is the circumcenter by definition. Hence we are done.

Start by drawing a nice big diagram.

The first thing we should do is to find out where $P$ can lie. Since it is indirectly suggested that $P$ can coincide with $I$, an immediate guess (if you know about the superlemma) is that $P$ might lie on the circle through $B,I$ and $C$. We can show that this is the case by exploiting the angle condition of the problem. The angle sum in $BPC$ gives that

$$\angle PBA + \angle PCA = \angle PBC + \angle PCB = 180^{\circ} - \angle BPC.$$

But we also have that

$$\angle BPC = 360^{\circ} - (\angle CPA + \angle APB) = 360^{\circ} - (360^{\circ} - \angle A - \angle PBA - \angle PCA)$$

from angle sum in quadrilateral $ABPC$. 

Putting these two expressions for $P$ together, we get that 

$$\angle BPC = \angle A + \angle PBA + \angle PCA  = \angle A + (180^{\circ} - \angle BPC) $$

which can be simplified to 

$$\angle BPC = \frac{180^{\circ}+\angle A}{2} = 90^{\circ} + \frac{1}{2} \angle A.$$

Since this angle is constant, $P$ must lie on a circle through $B$ and $C$ by the inverse of the inscribed angle theorem. Setting 

$\angle PBA = \angle PBC$ and $\angle PCA = \angle PCB$ we see that $P$ in this case is the intersection of two bisectors and thus the incenter. This implies that $I$ lies on this circle, and we have just confirmed that we are indeed working with the circle from the superlemma. We thus let $S$ be the superpoint and finish the problem by using the triangle inequality.

Using the triangle inequality on triangle $ASP$, we get that 

$$AP + PS \geq AS = AI + SI.$$

with inequality if and only if the triangle is just a line. Since $PS$ and $SI$ are both radii of the circle through $B,I$ and $C$, we are left with 

$$AP \geq AI.$$

Also, since the triangle is only a line when $P=I$, the equality only holds if and only if $P=I$ and we are done.

Remember how we found the connection between the inscribed circle and the angle bisectors? The same logic holds here.

The feet from the A-excenter to the lines $AB, BC$ and $CA$ must all have equal length because of the tangency. Since the altitudes from a point to two lines are only equal if the point lies on the bisector of the angle formed by the lines (remember this from the lesson about the incenter), the figure makes it clear that the A-excircle lies on the relevant bisectors.

Let the angle be $\angle BAC$ and let $D$ lie on the extention of the side $AB$ beyond $A$. Furthermore, let $P$ be a point on the interior angle bisector and $Q$ a point on the exterior one (these are just there so that we can work with the angle between the bisectors as $\angle PAQ$).

By the definition of the external angle bisector, it is in this case the internal angle bisector of angle $\angle DAC$. We then have

$$\angle PAQ = \angle PAC + \angle SAQ$$

$$= \frac{1}{2} \angle BAC + \frac{1}{2} \angle CAD = \frac{180^{\circ}}{2} = 90^{\circ}$$

and we are done.

From the previous exercise about the perpendicularity of the bisectors, we know that $\angle IBI_A$ and $\angle ICI_A$ both are $90^{\circ}$. We then have that opposite angles $\angle IBI_A + \angle ICI_A = 180^{\circ}$ which means that $I,B,I_A$ and $C$ all lie on a circle and we are done.

We already know that $HE$ is an angle bisector of $\angle DEF$. Since $CA$ is perpendicular to $HE$, $CA$ must then be the external angle bisector of $\angle DEF$. In a similar way, we see that $AB$ is an external angle bisector of $\angle EFD$. $A$ is then the intersection of two external and one internal angle bisector of the triangle $DEF$, and as such an excenter. That $B$ and $C$ are also excenters of $DEF$ can be shown similarly.

Since $AX, BY$ and $CZ$ are medians, the points $X,Y$ and $Z$ are of the midpoints of triangle $ABC$ as in the figure below.

Note that we have previously already shown $AZXY$ to be a parallelogram. This means $|XY| = |AZ|, |ZX| = |AY|$ and since $YZ$ is a common side of triangles $XYZ$ and $AZY$, these triangles are congruent by side-side-side. Similar calculations give the remaining two triangles.

It turns out that the ortocenter of triangle $XYZ$ is simply the center of the circumscribed circle of triangle $ABC$. 

To see that this is the case, simply notice that the configuration in the figure above is almost identical to the one in our proof of the existence of the orthocenter. We have just switched the roles of the triangles $ABC$ and $XYZ$. Because of this, we already know that the altitudes in triangle $XYZ$ are perpendicular bisectors of the sides of triangle $ABC$. This proves that the orthocenter of triangle $XYZ$ is the center of the circumscribed circle of triangle $ABC$.

We know that $AZY$ is similar to $ABC$, and that both have the exact same orientation. 

This means that the line $l$ through the orthocenter and centroid of triangle $AZY$ is parallel to $HG$. If we rotate $AZY$ $180^{\circ}$ about the midpoint of $ZY$, it lands on $XYZ$ and $l$ lands on $s$ since $AZY$ and $XYZ$ are congruent. Since a line rotated $180^{\circ}$ is parallel to the original line, this shows that $HG \parallel s$.

From our previous exercises we already know that the orthocenter and centroid of $XYZ$ are just $O$ and $G$. Therefore $OG$ and $s$ are the same line and the result follows.

Note that $HG$ and $OG$ are parallel. They also have a common point, $G$, and so they must be the same line and the result follows.

Simply notice that $GH$ and $OG$ are correspoinding segments in the similar triangles $ABC$ and $XYZ$. Since the scale factor between these triangles is $2:1$, we then know that $GH:OG = 2:1$.

The orthocenter, centroid and center of the circumscribed circle concur when the triangle is equilateral. Therefore it does not make sense to talk about an Euler line in this case, since there are infinitely many lines that go through a single point.

Let $G$ be the intersection of $OH$ and $AM$. Since we know that the centroid of triangle $ABC$ must lie on both the Euler line $OH$ and the median $AM$, $G$ must be the centroid.

We now use the ratios between the centers on the Euler line that we found. We have that $\frac{HG}{GO} = 2$, and we also know that $\frac{AG}{GM}=2$ from the ratios of the medians. This means that $\frac{HG}{OG}=\frac{AG}{MG}=2$ and since $\angle HGA = \angle OGM$, triangles $HGA$ and $OGM$ are similar by side-angle-side similarity. The ratio of similarity is $2:1$, which means that $HA$ is twice as long as $MO$. Because $N$ is the midpoint of $HA$, we then have that $|HN|=|MO|$.

We also know that $AH$ is an altitude and $MO$ is a perpendicular bisector of triangle $ABC$. This means that they are both perpendicular to $BC$ and thus parallel to each other.

Since $HN$ and $MO$ are both parallel and of equal length, $NHMO$ must be a parallelogram. This implies that $HM$ and $NO$ are parallel and we are done.

Let $D$, $E$ and $F$ be the midpoints of $BC$, $CA$ and $AB$. It is enough to show that

$$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1,$$

and since $D$, $E$ and $F$ are midpoints, we have that

$$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1 \cdot 1 \cdot 1 = 1.$$

Let $D$, $E$ and $F$ be the feet of the altitudes from $A$, $B$ and $C$ respectively. 

It is enough to show that 

$$\frac{\sin{\angle BAD}}{\sin{\angle DAC}} \cdot \frac{\sin{\angle CBE}}{\sin{\angle EBA}} \cdot \frac{\sin{\angle ACF}}{\sin{\angle FCB}} = 1.$$

But because of the right triangles we have that the left hand side is equal to

$$\frac{\sin{90^{\circ} - \angle B}}{\sin{90^{\circ} - \angle C}} \cdot \frac{\sin{90^{\circ} - \angle C}}{\sin{90^{\circ} - \angle A}} \cdot \frac{\sin{90^{\circ} - \angle A}}{\sin{90^{\circ} - \angle B}}$$

which is equal to $1$.

Because of the midpoints, we know that $BD' = DC$ and that $D'C = BD$, so we get that 

$$\frac{BD'}{D'C} = \frac{DC}{BD}$$

We similarly get that

$$\frac{CE'}{E'A} = \frac{EA}{CE}, \; \frac{AF'}{F'B} = \frac{FB}{AF}.$$

Combining all expressions, we get that

$$\frac{BD'}{D'C} \cdot \frac{CE'}{E'A} \cdot \frac{AF'}{F'B} = \frac{DC}{BD} \cdot \frac{EA}{CE} \cdot \frac{FB}{AF},$$

and since $AD$, $BE$ and $CF$ are concurrent we know by Ceva's theorem that this is equal to $1$. By Ceva's theorem we are then done.

Let $AD'$ be the reflection of $AD$ over the angle bisector of $\angle A$. 

Then we know from the symmetry that

$\angle BAD' = \angle DAC$ and $\angle D'AC = \angle BAD$. This means that 

$$\frac{\sin{\angle BAD'}}{\sin{\angle D'AC}} = \frac{\sin{\angle DAC}}{\sin{\angle BAD}}.$$

We get similar expressions for the other two lines, and since the original $D$, $E$ and $F$ angles cancel out to $1$, so does the new angles also. By Ceva's theorem we are then done.