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.
LetŐs start from the middle of segment BC.
Barney will hit two walls before getting back because when we draw parallels from our starting point T, we make a triangle RST similar to triangle ABC, and 4 congruent triangles are constructed.
Let Barney start at a point when he is (1/3) of the distance of BC from B, call this point T.
We get nine triangles that are all congruent and all similar to triangle ABC. So perhaps it is 32 and 22 instead of 2*2 and 3*2.
Based on the previous two that we have y = n2 where n is the number of segments that BC is broken down into, and y is the number of triangles formed.
So, let Barney start at a distance that is a (1/4) of BC from point B, call this point V.
Our conjecture does not hold.
There are 7 triangles and 3 trapezoids.
Perimeter of ABC = Perimeter of the dashed lines
The area of the smaller triangles have the sides of the same length => areas are the same.
Apparently, the areas of the triangles are equal and they are always equal.
Let Barney be outside the room at point P. If he starts walking, he will just keep going on forever or until he gets tired.
So, in this case Barney will never be inside the room.
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.
1. Explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P.
2. Conjecture? Prove it! (you may need draw some parallel lines to produce some similar triangles, OR, you made need to consider areas of triangles within the figure) Also, it probably helps to consider the ratio
3. In your write-up, after the proof, you might want to discuss how this theorem could be used for proving concurrency of the medians (if P is the centroid), the lines of the altitudes (if P is the orthocenter), the bisectors of the angles (if P is the incenter), or the perpendicular bisectors of the sides (if P is the circumcenter). Concurrency of other special points?
4. Explore a generalization of the result (using lines rather than segments to construct ABC) so that point P can be outside the triangle. Show a working GSP sketch.
To explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P, click on the link below:
Clearly, each product in the different triangles is the same, hereŐs why:
By CevaŐs Theorem, lines AD, BE, and CF are concurrent if and only if
So construction gives us:
Because DBPD and DCPD have the same height, we have
It follows that,
From the two equations for BD/CD, we get:
BD/CD = Area(APB)/Area(APC)
If we had started with the other two sides instead of BC, we would have gotten the following:
CE/AE = Area(CPB)/Area(APB)
AF/BF = Area(APC)/Area(CPB)
Multiplying these three equations gives us:
Since the cevian is the line segment joining a vertex of a triangle to any given point on the opposite side, it is convenient to apply the CevaŐs Theorem for proving the concurrency.
This theorem is useful to prove of the concurrency of the medians (if P is the centroid), the lines of the altitudes (if P is the orthocenter), and the bisectors of the angles (if P is the incenter) or other special point P.
Besides showing the concurrency of segments, we also have AF = EA, FB = BD, DC = CE.
Say, for instance p is outside of the triangle, click below to see a sketch of this:
Seemingly, our property holds even if p is outside of the triangle.