Let Us Consider Yet another Triangle . . .

Final Project

by

Robin Kirkham

 


Part A

Consider yet another triangle (we have not considered enough triangles) ABC. Select a point P inside the triangle and observe the lines segments AP, BP, and CP extended to an intersection with the opposites sides at points D, E, and F.

 

Let us explore further the segments (AF)(BD)(EC) and (FB)(DC)(EA) for a variety of triangles and change the location of point P.

 


 

 

This exploration of the segments of each of the side of the triangle

(no matter the length - as changed by the position of P and the movement of the vertex)

the cooresponding segments when multiplied produced the same values in the same areas.


Part B

 

It is not enough to look at the simple triangle and stop there (that is no fun).

Let us consider the ratio of

(AF)(BD)(CE)/(BF)(CD)(AE)

According to our calculations, this ratio equals 1 everytime (even when P or a vertex is move thus changing the length of the segments).

Given: triangle ABC, with -

F on line AB,

E on line AC,

D on line BC

line AD, line BE, and line CF are concurrent at P

 

Proof or Conjecture?

 

Prove:

Line XY is constructed to be parallel to line BC below.

Connect vertices C and B through point P making lines CX (line through P) and BY (line through P)

 

On the above diagram make the following observations:

Dreaw a line through A such that it is parallel to line BC meeting line CP at point X and line BP at point Y.

Notice the two triangles in yellow AEY - CEB

therefore, (1.)

 

for triangles BFC-AFX therefore, (2.)

 

 

for triangles CDP-YAP therefore, (3.)

 

for triangles BDP-XAP therefore, (4.)

 

 

from the ratios (3.) and (4.) we get

(5.)

 

Now if we multiply (1.), (2.), and (5.) providing the followowing:

and then you can show the following:

 


 

Part C

show that when P is inside triangle ABC, the ratio of the areas of triangle ABC and triangle DEF is always greater than or equal to 4.

When is it equal to 4?

In the above triangle ABJ and DFK areas are compared. Notice that the ratio of the two areas is greater than 4.

Below the point P has been adjusted and the two areas are compared and again the ratios is still greater than 4.

 

Furthermore, the third diagram (above) has moved point P yet again to the centroid and still the areas are compared and are a ratio of 4. No matter what location point P is at within the triangle the ratio of the two areas remains at a minimum of 4.

 

 

Furthermore, the third diagram (above) has moved point P yet again to the centroid and still the areas are compared and are a ratio of 4.

Conclusion:

No matter what location point P is at within the triangle the ratio of the two areas remains at a minimum of 4.

As we observed in assignment 4 (centers of a triangle) the CENTROID of a triangle is the common intersection of the three medians.

This makes the areas of the six triangles (BFP, FAP, AKP, KJP, and JDP) equal.

 

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