You have 1:15 hrs. to complete this exam. Each question is worth 20 points. Please show all work on these sheets. This is a closed book, closed note exam. Good Luck !!!
1. Define the following italicized terms by completing the sentences below. (Your definitions need not be verbatim reproductions of the book or class notes, but they must be correct!)
a) An excenter of a triangle is ...the intersection of two external angle bisectors.
b) The Euler line of a triangle is the line segment determined by ...the orthocenter (H) and the circumcenter (G).
c) A median of a triangle is ...a line segment drawn from a vertex to the midpoint of the opposite side.
d) A constructible number is the length of a line segment which ...can be obtained from a unit length by straightedge and compass constructions.
e) The 9-point circle of a triangle contains the following 9 points of a triangle : the three midpoints of the sides of the triangle, the three feet of the altitudes of the triangle and the three midpoints of the segments drawn from the vertices to the orthocenter of the triangle.
2. a) To construct the Euler line of a right triangle you only need to construct one point and then join this point to another to obtain the line. What is the point you need to construct and how would you use a straightedge and compass to construct it?
In a right triangle the orthocenter is the vertex at the right angle, so you need only construct the circumcenter to obtain the Euler line. In a right triangle, the circumcenter is the midpoint of the hypotenuse, since the hypotenuse will be a diameter of the circumcircle. Use the straightedge and compass to find the midpoint of the hypotenuse by describing arcs with the same radius and centers at the endpoints of the hypotenuse. Join the points where these arcs intersect with the straightedge and this line intersects the hypotenuse at its midpoint.
b)In the construction of a triangle given the measure of one (acute -oops, my mistake for leaving this out) angle, the length of the internal bisector of that angle and the length of one adjacent side of that angle it may happen that the construction fails. Carefully describe the one situation in which this can occur.
Since the angle is given, we can construct the angle bisector and knowing its length we can find the point on it which lies on the third side of the triangle. Since we also know the length of a side adjacent to the angle, we obtain a second point on the third side of the triangle. Joining these two points would give the third side of the triangle. This construction will fail if this third side does not intersect the other side of the angle above the angle bisector. In turn, this will occur if the length of the angle bisector is too long in comparison with the given side of the angle.
3. Given triangle ABC with the line segments measured as below, what can you say about the collinearity of points E, F and G and the concurrency of lines a,b and c. State the names of any theorems you use to reach your conclusions. NOTE: The line segments other than the sides of the triangle may not be accurately drawn.
Consider (AF/FB)(BE/EC)(CG/GA) = (3/1)(7/-12)(4/1) = -7. Since this is not -1, Menelaus' theorem says that the points E, F and G are not collinear. Letting the points where the lines a,b and c intersect the sides of the triangle be labeled by A', B' and C' respectively, we examine (AC'/C'B)(BA'/A'C)(CB'/B'A) = (2/2)(2/3)(3/2) = 1. By Ceva's theorem, the lines a, b and c are concurrent (meet at a point).
4. Choose one of the following three problems.a. Prove that the radius of the 9-point circle of a triangle is half the radius of the circumcircle of that triangle.
In the diagram below, with H the orthocenter, O the center of the circumscribed circle, N the center of the nine-points circle and J the midpoint of AH (on the nine-points circle), consider the triangle AHO. Since J is the midpoint of AH and N is the midpoint of HO, the segment JN is parallel to and half the length of the side OA. But JN is a radius of the nine-points circle and OA is a radius of the circumcircle.
b. Prove that the segments from the vertices of a triangle to the points of tangency of the incircle with opposite sides of a triangle are concurrent (meet at a point called the Gergonne point).
Let A', B' and C' be the points of tangency opposite to the vertices A, B and C of the triangle. Since the sides of the triangle are tangents to the incircle, AB' = AC', BC' = BA' and CA' = CB'. Thus, (AB'/B'C)(CA'/A'B)(BC'/C'A) = 1. So, by the converse of Ceva's theorem, the lines are concurrent.
c. In an arbitrary triangle ABC, let D be the foot of the altitude drawn from A to BC and let A', B' and C' be the midpoints of the sides of the triangle. Prove that DB'C'A' is an isosceles trapezoid.
Since B' and C' are midpoints of the sides of the triangle, the line joining them is parallel to the third side of the triangle, which includes the line segment DA'. Thus, the opposite sides, B'C' and DA' of this quadrilateral are parallel, so it is a trapezoid. Since, A' and C' are midpoints of the sides of the triangle, A'C' has length equal to 1/2 the length of AC. In the right triangle, ADC, the line segment DB' joins the right angle vertex to the midpoint of the hypotenuse. The length of such a line is 1/2 of the hypotenuse, which is AC. Thus, DB' and A'C' have the same length, so the trapezoid is isosceles.
5. Write a paragraph which expresses your understanding of the angle trisection problem. Include in this paragraph specific statements dealing with the following points.
Give an accurate statement of the problem.
What is the mathematical status of the problem?
How important is this problem in the area of geometric constructions?
How would you respond to someone who insisted that they had an angle trisection construction?
The angle trisection problem is to trisect an arbitrary angle using only a compass and a straightedge. It can be proved that this is impossible to do, however, angles can be trisected if additional tools are used (for example, using a straightedge that has two marks on it). The problem is of some theoretical interest because it shows that there are limitations to what can be done using only the two traditional tools, however it is of no practical importance since angles can be trisected by other means. If someone insisted that they had an angle trisection construction, I would direct them to Underwood Dudley's book, A Budget of Trisections, hoping that they would understand the message contained in that book.