Bouncing Barney and Ceva's Theorem

Final Assignment

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A.  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.

Here is a GSP file of Bouncing Barney (triangle with line segments).  A script tool has been included in the download.  Clickhere.

If you want to look Barney’s exterior path (triangle with lines), click here for a GSP file with this script tool.

Let us begin the proof that Barney will eventually return to his starting point by labeling the triangle like shown below:

Algebraically, line segment AB is the line y = a/b x, line segment AC is the line y = b/(a-c) x – bc/(a-c), and line segment BC is the line y = 0.  I will not go through all the algebra but the first few steps.  First, construct a line through (d,0), where Barney starts, parallel to AC and intersecting AB.  This line would have the same slope as AC and goes through (d, 0).  So the equation of this line would be y = b/(a-c) (x-d).  This line would intersect AB at the point (ad/c, bd/c).  This is the first leg of Barney’s journey.  The proof consists of 5 more equations with intersections on the sides of the triangle as shown below:

Barney turns 5 times with 6 legs back to the starting point.

What about choosing a point exterior of the triangle?

Notice the Superman shield?  Also, note that Barney returns to his starting point with 5 turns and 6 legs of the journey around the exterior of the triangle.

Barney is located at (d, 0).  The line segment formed from the starting point (d, 0) to the intersection of line AC has an equation y =  b/a (x-d) since it is parallel to line AB.  Thus the intersection point is

(a + c – ad/c, (abcabd)/c(a – c)).  Look familiar?  Similarly to the above algebraic proof, we can conclude that Barney does make it back to (d, 0).

Conjecture:  If Barney starts at a midpoint of one of the sides of the triangle, he will have 2 turns to make and 3 legs of the journey to make it back to his starting point.  Also, several similar triangles are formed.  Look at the picture below.

Conclusions:

1.      Barney returns to his starting point if walking inside or outside the triangle.

2.      If Barney starts at the midpoint, he returns after 2 turns to his starting point.

3.      If Barney starts at any other point on BC besides the midpoint, he returns after 5 turns to his starting point.

4.      Perimeter:

a.      If Barney starts on a midpoint of the side of the triangle, he will travel half of the triangle’s perimeter.

b.      If Barney starts on any other point besides the midpoint of a side of the triangle, he will travel the triangle’s perimeter before returning to his starting point.

These conclusions could be proved using properties of similar triangles (unlike the algebraic proofs above).

B. Ceva's Theorem. Consider any triangle ABC. Select a point P inside the triangle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively.

Exploration:

Consider the ratio below in any triangle:

The above ratio is equal to 1 if inside the triangle ABC.

For this proof, draw a line through A parallel to line segment BC as shown below.

Constructing the line segments above, several similar triangles are formed.

Δ AET ~ Δ CEB, so AE / EC = AT / CB.  (I)

Δ BFC ~ Δ AFS, so BF / FA = CB / SA.  (II)

Δ CDP ~ Δ SAP, so CD / SA = DP / AP.  (III)

Δ BDP ~ Δ TAP, so BD / TA = DP / AP.  (IV)

Looking at (III) and (IV), CD / SA = BD / TA.  With a little algebra, this proportion is the same as CD / BD = SA / TA.  (V)

By multiplying (I), (II), (V), AE / EC * BF / FA * CD / DB = AT / CB * CB / SA * SA / TA = 1, or

Conjectures:

1.      The medians of any triangle are concurrent.

2.      The altitudes of any triangle are concurrent.

3.      The interior angle bisectors of a triangle are concurrent.

Let us look again at the centroid case:

In triangle ABC, line segments AD, BE, and CF are medians.  Therefore, AF = FB, BD = DC, and CE = EA.

So (AF)(BD)(CE) = (FB)(DC)(EA), or

Thus, according to Ceva, line segments AD, BE, and CF are concurrent.

Exploration of Ceva’s Theorem using lines rather than segments to construct ABC so that point P can be outside the triangle.   Below are sketches of when the points P, D, E, and F are collinear.

According to Menelaus, if we were to take direction into account in the two figures above then the ratio would equal -1 instead of 1.

Click here to open a GSP file with the collinear cases above.