Remind Pat about list of things: meeting Monday of NCTM, consider Sit. 46 the previous
version or marked-up version, expectation of work during SB,
Class Notes
EMAT 6600
March 7, 2007
Problem (as printed by Jim Wilson): Grass growing. Hint: there will be an even number
of oxen.
More Problems
 (Starting with incircle and inscribed tangent circle from Sunday)
 Start with a circle and 6 circles that are tangent to it. Connect the opposite points of
tangency. The segments are concurrent, meaning they share a point in common for
all three segments. It does not matter how the 6 circles are tangent to the original
triangle, the theorem still holds. Problem was supposedly purposed around 1800 at a
temple in Japan. This is an extension from Sunday’s problem. (The three sides of the
triangle are circles of infinite radius.) Could this help us to understand how to solve
the problem with the previous problem about the triangle, incircle, tangent circles,
and corresponding radii?
 Start with a triangle. Extended the sides of the triangle. Construct the incircle and
the three excircles. The center for all four circles will be on the angle bisectors of the
vertices of the triangle. There is a right angle that is formed by the connection of the
angle bisector of the triangle and one of the external angles of the triangle as well.
The three excircles’ centers will construct a circle that is circumscribed by the
original triangle. We want to prove that the reciprocal of the radius of the incircle
equals the sum of the reciprocals of the radii of the excircles. (Mathworld suggestion:
Consider the altitude of the excircle. This is not a true statement.) Look at the line
segments formed by the center of the incircle and the vertices of the triangle formed
by the centers of the excircle. Take the circumcircle of the original given triangle.
Show the relationship between the circumcircle and the segment connecting the
incenter and each of the excenters.
 “Warm-up” problem: Start with a right triangle. Construct the altitude from the right
angle through the hypotneuse. What is the relationship between the length of the
altitude and the legs of the triangle? Prove that the reciprocal of the height of the
altitude squared equals the sum of the reciprocals of the two legs squared. (My
solution is written out on the back of the grass and cow problem.) Variation: Given a
semi-circle and a point on the diameter that is not the center. The line perpendicular
to the diameter intersects the semi-circle at a point, the segment formed by these two
points as a length of the geometric mean of the two portions of the diameter. There is
a segment that represents the radius of the semi-circle. Try to find the relationship
between these two segments.
 Start with a right triangle and pick an arbitrary point on one of the legs. Construct a
perpendicular line through the arbitrary point. Form a segment based on the
intersection of the line along with the sides of the triangle. Using the GSP sketch
prove that XY = TU + D2
 Start with a cyclic quadrilateral. Construct the diagonals of the quadrilateral. Look at
four triangles that the quadrilateral creates. Using the associated GSP sketch,
construct the incircles for the following triangles: ABC, BCD, CDA, and DAB.
Remind Pat about list of things: meeting Monday of NCTM, consider Sit. 46 the previous
version or marked-up version, expectation of work during SB,
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Connect the incenters. Prove that the centers of the incircles are the vertices of a
rectangle. However, the order of the vertices must be maintained for this property to
hold. Suggestion: Look into the angle bisectors that formed the incircles.
“Warm-up” problem: Start with a circle and a diameter. Divide the diameter into 2
pieces. Make each segment the diameter of a circle. Question: How does the
circumference of the original circle compare to the circumferences of the two smaller
circles? Could the same argument be developed using similarity?
Three Circles Problem. Also look at the variations of the problem. (Hint: one of the
variation problems will have something to do with circles.)
Start with any triangle. Construct a cevian. Construct the incircles of the two
triangles created on the interior of the original triangle. Find the one point where the
two circles have the same area. Additionally, prove that the angle formed by the
incenters and the point of intersection of the cevian and side is a right angle.
(Chances are this proof would have no bearing on the proof of circles with equal
area.)
Construct a circle and a circle whose center is a point on the original circle.
Question: At what placement of the circles is the area of the intersection of the two
circle exactly half of the original circle? (This intersection is called a lune.)
Construct a circle and divide the interior region into fourths. Connect the points of
intersection to create a line segment. Let the line segment be the diameter of an arc.
Prove that it is possible to square a lune. In fact, there are five lunes which can be
squared. Follow-up: any polygon can be constructed so that the area of the polygon is
the exact same as the area of a corresponding square. (The original problem is called
the quadrature of the lune.)
Problem from the website: (sine of 18 degrees)^2 + (cosine of 36 degrees)^2 = ¾.
Look at the golden ratio. We know that phi^2 = phi + 1. How could we find phi^3?
We could re-write phi^3 = phi^2*phi = (phi +1)*phi = phi^2 + phi = phi + 1 + phi =
2*phi + 1. What would happen if we continue to phi^4 = (phi^3) * (phi) = (2*phi +
1)*phi = 2phi^2 + phi = 2(phi + 1) + phi = 2*phi + 2 + phi =3*phi + 2. Moving on to
(phi)^5 = 5*phi + 3 (after the calculations are performed). The same calculations
could be performed on the negative integer exponent. Extension: Look at the
Fibonacci sequence. The infinite of consecutive terms of the Fibonacci sequence
equals phi. What happens if we compare terms that are two terms apart in the
Fibonacci sequence? Turns out the infinite limit of this comparison/ratio is phi
squared. What about three terms apart? Answer: phi cubed.