Time for MATH MINUTE! (provide your favorite theme music
here).
Get out
your paper & pencil, because I have a new puzzle for you!!
Hints/Responses: PUZZLE #89 Triangles in triangles
A. Make a triangle
ABC and choose a point P in the interior of the triangle.
Now
draw lines through P parallel to the sides of the triangle.
These
3 lines will form 3 triangles with the sides of triangle ABC as follows.
Did
you notice that these 3 triangles look like triangle ABC?
Are
they similar to ABC (have the same angles as ABC)?
Why?
Since the sides of
each triangle are parallel to one of the sides of triangle ABC their angles
will automatically be equal to one of the angles of triangle ABC.
This
is actually a result due to Euclid in Book I of his The Elements (see Prop. 29).
Basically what it says is that if you have a pair of parallel lines and you
have a line falling on them, then the exterior angle is equal to the
interior (between the parallel lines) and opposite angle.
For
example, in our situation we can say that angle EDP equals angle A, since side
AB and line DPI are parallel, and line AC falls on them, so exterior
angle EDP is equal to the interior and opposite angle A. Similar arguments hold
for all of the angles of the triangles inside triangle ABC.
What
about the other white shapes inside ABC. What geometric shapes are they?
Why?
The white shapes are
parallelograms, since their opposite sides are parallel.
B. If you
make different choices for point P you get different sizes for the 3 triangles,
unless you choose P near the geometric center (centroid*) O of ABC (see figure
below).
At
the centroid point O the 3 triangles appear to be congruent and therefore equal
in area.
Are
they?
Why?
Since we are given
below the result that the centroid divides the medians in the ratio of 2 to 1
(see *), we know that the side DE is the middle 1/3 of side AC, similarly for
GF and HI. Also, the point P=O is now the center of lines DI, GH and EF (why?).
This
means that the similar triangles DEP, HIP, and FGP are in fact equal
(congruent).
BONUS: The 3 triangles inside ABC change sizes as you
move the point P. Is there a point where the sum of the areas of these 3
triangles is a maximum or a minimum?
Why?
Yes, the minimum area
is when P is at the centroid point O, and the maximum area is when P is at any
of the vertices.
When
you choose P close to a vertex, say C, the triangle opposite P from the vertex,
FGP in this case, is very large in area
and approaches the size of ABC when P approaches the vertex.
On the other hand,
when you choose P at O, the sum of the triangle areas is a minimum. This is
easy to see, but not so easy to prove.
Basically,
when you move the point P away from O, at least one of the triangles begins to
increase in area faster than the area of the remaining triangle(s).
See
if you can find a simple argument for this result.
Joel
Foisy (foisyjs@potsdam.edu) has
constructed a nice argument which uses multivariate calculus.
* This is the point where the medians (the lines connecting each
of the vertices with the midpoint of the opposite side) intersect. In physics
its the point where the object has its center of gravity or mass. This point
divides the medians in the ratio of 2 to 1.
BTW this is a simple
result to show. Draw a picture of the triangle ABC with 2 of the medians, say
AT and BS, as above. Then connect S and T, and recall that the line connecting
the midpoints of two sides of a triangle is of length ½ the length of
the 3rd side. Now compare triangles STO and ABO.
Have
fun!
Send
your comments, ideas and solutions before Monday to the email below, and
in the subject line be sure to put
MM in the subject line
Visit
us here online at:
http://www2.potsdam.edu/parksjm/MM1.1.htm
to see the results every Friday.
See you next time on MATH MINUTE! (theme music fades out here).