I walked into to school and everything was different. It was like everything that I thought was there was gone, but that’s ok because I am a quick learner. Unlike some people, I thought to myself. As I walk down the tremendously lengthy hallway I saw a bunch of people talking with their friends. I don’t have any friends, but that’s ok because I have my books. There he goes, the most popular guy in school, or ­­­­­swagalishious, as he calls it. I don’t know how he got there, being a bully and putting everyone else below you shouldn’t boost up your popularity. I also don’t have any calculations that would plug in either. I walk into the classroom and sit right in the front, like usual, and pull out my books. I can usually get away with reading my high school textbook in class and college textbooks after classes.

 

A loud noise runs all around the school, which means my favorite time of day is here, when class starts. The teacher walks through the door and says pop quiz. My face lights up with joy and I grab my fully sharpened to the point #2 pencil. He gives my quiz and I can’t wait to start.

 

Question 1:

Given a triangle ABC. BL is the bisector of angle ABC, H is the orthocenter and       P is the mid-point of AC. PH intersects BL at Q. If ∠ABC=β, find the               ratio PQ:HQ.If QR⊥BC and QS⊥AB, prove that the orthocenter lies on RS.

 

Duh, this is sooooo easy, I thought and quickly wrote down the answer.

Answer 1:

In the figures below, I have added the circumcenter, U, and the centroid, E. I        have also placed Lon the circumcircle.

Note that since both are perpendicular to AC¯¯¯¯¯¯¯¯, we                                               have BH¯¯¯¯¯¯¯¯||UP¯¯¯¯¯¯¯¯; furthermore, |BH¯¯¯¯¯¯¯¯|=2|UP¯¯¯¯¯¯¯¯|. The        latter is because △PUE is similar to △BHE and

P=A+C2 and E=A+B+C3(1)

so that

P−E=A−2B+C6 and E−B=A−2B+C3(2)

Thus,

|UP¯¯¯¯¯¯¯¯|=Rcos(B) and |BH¯¯¯¯¯¯¯¯|=2Rcos(B)(3)

where R is the circumradius of △ABC.

Since the line containing UP¯¯¯¯¯¯¯¯ is the perpendicular bisector                              of AC¯¯¯¯¯¯¯¯, the point at which UP−→−−intersects the circumcircle                         of △ABC splits the arc between A and C in half. Of course, the bisector                      of ∠ABC also splits the arc       between A and C in half. Thus, the                                perpindicular  bisector of AC¯¯¯¯¯¯¯¯ and the       bisector of ∠ABC meet on the       circumcircle at L.

Note that △BHQ is similar to △LPQ. Equation (3) gives                                                   that |UP¯¯¯¯¯¯¯¯|=Rcos(B) so that

|PL¯¯¯¯¯¯¯|=R(1−cos(B))(4)

Therefore, (3) and (4) yield

|HQ¯¯¯¯¯¯¯¯|/|PQ¯¯¯¯¯¯¯¯|=|BQ¯¯¯¯¯¯¯¯|/|LQ¯¯¯¯¯¯¯|=|HB¯¯¯¯¯¯¯¯|/|PL¯¯¯¯¯¯¯|      =2cos(B)1−cos(B)(5)

which answers the first part.

Because △BUL is isosceles with central angle 2A+B=π−(C−A), we have

|BL¯¯¯¯¯¯¯|=2Rsin(A+B2)=2Rcos(C−A2)(6)

Equation (5) yields that |BQ¯¯¯¯¯¯¯¯|/|BL¯¯¯¯¯¯¯|=2cos(B)1+cos(B).                           Thus, (6) gives

|BQ¯¯¯¯¯¯¯¯|=2Rcos(C−A2)2cos(B)1+cos(B)(7)

Let X be the intersection of BQ¯¯¯¯¯¯¯¯ and RS¯¯¯¯¯¯¯. Since X is on the angle          bisector of ∠ABC, RS¯¯¯¯¯¯¯ is perpendicular                                                                  to BQ¯¯¯¯¯¯¯¯ and |BR¯¯¯¯¯¯¯¯|=|BS¯¯¯¯¯¯¯|.                                                                     Thus, |BR¯¯¯¯¯¯¯¯|/|BQ¯¯¯¯¯¯¯¯|=|BX¯¯¯¯¯¯¯¯|/|BR¯¯¯¯¯¯¯¯|=cos(B/2).                       Therefore,

|BX¯¯¯¯¯¯¯¯||BQ¯¯¯¯¯¯¯¯|=cos2(B/2)=1+cos(B)2(8)

Equations (7) and (8) yield

|BX¯¯¯¯¯¯¯¯|=2Rcos(C−A2)cos(B)(9)

Since ∠HBC=π2−C and ∠QBC=B2 we get that ∠HBQ=C−A2. Using (3), the                   orthogonal projection of BH¯¯¯¯¯¯¯¯ onto BQ¯¯¯¯¯¯¯¯ has length                                   is 2Rcos(B)cos(C−A2). Thus, the orthogonal projection of H onto BQ¯¯¯¯¯¯¯¯ is X.      Therefore, H lies on RS¯¯¯¯¯¯¯.

 

DONE!

Wow, easy 100%, and back to my textbooks. It has only been 5 minutes so I am going to get started on month 6’s homework.

 

The air around me whips me across the face as the doors to Chick-fil-a open cautiously. It only takes a few small steps before you’re flooded with technology, after all it is the year 2215! Apple is the main technological establishment due to most of the other companies joining on to Apple. Chick-fil-a is now only a small, yet nice building due to the lazier common delivery system. As you walk small, cute androids fly around you with the menu. They do not all swarm each person that walks through the door, instead they scan your mind and try to see what you would like. Being a more frequent customer they know that I don’t need one at all.

A few steps later, I reach the belt. The belt is where your order comes out. When I step onto the platform where it scans your brain to know what you want to order, it says in a very friendly voice “Thank you for choosing Chick-fil-a your order will be right out.” As it finishes up its sentence it spits out your food, and says the Chick-fil-a catch phrase… “My pleasure”. Since the technology is so advanced it knows my order and it is already paid for. A chair floats up to me efficiently and takes me to a table. When I am done eating it takes me to the door and brings me to my car which is waiting outside.