Bouncing barney

by

Hyeshin Choi

Barney is in the triangular room shown here. He walks from a point on BC parallel to AC. When he reaches AB, he turns and walks parallel to BC. When he reaches AC, he turns and walks parallel to AB. Prove that Barney will eventually return to his starting point. How many times will Barney reach a wall before returning to his starting point? Explore and discuss for various starting points on line BC, including points exterior to segment BC. Discuss and prove any mathematical conjectures you find in the situation.

First I will begin to consruct barney's path. Click here forGSP.

Case I. Barney starts from vertex. If Barney starts at a vertex of the triangle, then he walks the perometer of the triangle.Click here for GSP.

CASE II. Now Barney starts at the midpoint of the triangle. He returns the starting point since we know that ab which is parallel AB must meet the midpoint of BC and AC. Similarily, line ca meets the midpoint of AB and BC. Therefore Barney returns to the starting point. And the distance he walks is ab= ½AB, bc = ½BC,and ca= ½CA. Therefore he travels ab+bc+ca = ½(AB + BC + CA) .Click here for GSP.

CasaIII. Barney starts at any randompoint not on vertex nor a midpoint of the triangle.Click here for GSP. Does Barney returns to the starting point if he starts any point other than midpoint or vertex? I will prove this by assuming that S" is not equals to S.

Since Barney walks parallrel to each side of triangle ABC, this means that ES is parallrel to AC and HI is parallel to BC. Therefore, angle ACB = angle ES'B by the corresponding angle.

Now we have three congruent parallelograms which are EFCS, HIS'B, and HICG. Since the opposite sides are congruent, FC≅ES, HI≅BS, HI≅GC. Thus by side by side theorem, triangleFCG and ES'B are congruent. Therefore, S'=S which means that Barney returns to the starting point.

Case IV. Barney starts a points outside of the triangle. Click here for GSP.

Since MO||AB and ON||AC, angle BAC≅MON

Since MO||AB and MN||BC, angle ABC≅OMN

Since MN||BC and ON||AC, angle ACB≅ONM

Hence triangle ABC and triangle OMN are similar.

And BArney's path LGHIJK is on triangle OMN which means that Barney returns to the starting point.

Butterfly Theorem

by

Hyeshin Choi

Through the midpoint C of any chord AB of a circle, any chords ED and FG are drawn; lines FD and EG meet AB at points M and N. Given C is a midpoint of AB.

DE and FG passing through C

FD and EG meet AB in M and N

Prove C is the midpoint of AB.

First, I will begin to construct a butterfly. Click here to see GSP.

1. We know that the angle DFG= angle GED by the inscribed angles.

2. Triangle FCD ans ECG are similar since FD/FE=EG/EC

3. Construct perpendicular from O to DF and from O to EG like below. Click here to see GSP. Then FD=2FH and EG=2EJ.

4. Ln(2): FH/FC=EJ/EC but angle DFG= angle GED. Therefore triangle FCH and triangle ECJ are similar.

5. Now construct line CH and CJ. Then by the corresponding angles, angle FHC and EJC are congruent. Click here for GSP.

6. Construct a segment OC. Then construct MO and NO. Then OCMH is cyclic ( angle MHO+angle OCM=180 degree)

Therefore angle MHC=angle MOC.

Similarily OCNJ is cyclic. Therefore angle NJC = angle NOC.Hence angle MOC=angle NOC.

7. Finally triangle OCM is equal to triangle OCN by angle-side-angle. Therefore we can say that C is the midpoint of MN.