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.