I have investigated two problems for this assignment the first one is:

# 11. Construct any acute triangle ABC and its circumcircle. Construct the three altitudes AD, BE, and CF. Extend each altitude to its intersection with the circumcircle at corresponding points P, Q, and R.

Find

,

Observations:

1) look at the triangle HPC, call the measure of the interior angles as x, and y. We will have 2y as the measure of the arc that corresponds to the angle DCP.

2) Then the measure of BAP angle will be "y" because of the same reason; since AB ray and PA ray defines the corresponding arc on the circle.

3) If we look at the two triangles; FBC and DBA we will see that B angles are same and both F and D angles are right angle (since they are altitudes rays from C and A vertexes of the triangle). The sum of any triangles 'interior angels are 180 degrees, x has to be equal to y.

4) So we will have two triangles HDC and PDC which are similar (because the corresponding angles are same). And actually, the side DC is common for both which makes these two triangles equal, where all the corresponding sides are equal. We will have HD length being equal to DP.

4) We can use the same reasoning to see the length of EH and EQ ; and RE correspondingly FH are equal.

5)We can get some relationships between the orthogonal ray segments that cuts the circle and the smaller pieces of that orthogonal segments.We will basically use the informations that we derive in 1) through 4), related to having equal measure of length for the pieces such as HD and DP .

6)We were asked to find

So, by 5)

,

Then multiplying each length (which are actually the altitudes of the triangles) with the corresponding bases we will have the areas,

While having the whole areas for the denominators of the equation we will have the below areas of small triangles; which makes a whole ABC triangle area when added up.

,

,

Where

,

So,

and we will get

1. Construct any triangle ABC.

2. Construct the Orthocenter H of triangle ABC.

3. Construct the Orthocenter of triangle HBC.

The Orthocenter of HBC is at the same time, vertex of A of ABC triangle.

4. Construct the Orthocenter of triangle HAB.

The Orthocenter of HAB is vertex C of triangle HBA.

5. Construct the Orthocenter of triangle HAC.

6. Construct the Circumcircles of triangles ABC, HBC, HAB, and HAC.

My conjecture is if we connect the points (the centers of the circumcircles 2 by 2) we will have the same (area, perimeter, same interior angles) ABC, HAB, HAC and HBC triangles upside-down.

The other possibility is having a "cube" by connecting all the intersection points and the centers of the circumcircles.

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