1. A triangle has how many:
Ans: 1, 1, an infinite number.
4. Give the maximum and minimum number of points a triangle and its nine-point circle might have in common.
Ans: 6, 3 Solution:A circle can intersect a side of a triangle in at most two points. If this occurs for all three sides, there will be 6 points in common (maximum). Since the nine-point circle must pass through the feet of the altitudes, and the midpoints of the sides, if these coincide (i.e., in an equilateral triangle), then the nine-point circle will only intersect the sides of the triangle in one point, giving 3 points in common (minimum)
7. In the proof of Thm 4.16, prove that DB'C'A' is an isosceles trapezoid.
Solution: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.
9. Complete the details of the proof of Thm. 4.17 not included in the text.
Solution:The two details missing in the proof are that AHBI is a parallelogram and that AG = 2GA'. This second fact follows from the fact that the centroid trisects each of the angle bisectors that pass through it. In the diagram (Fig. 4.19), CI is a diameter of the circle. This means that angles CAI and CBI are right angles. Thus, IB is parallel to altitude AD (hence AH), and IA is parallel to altitude BH. Therefore, AHBI is a parallelogram.
15.
Solution: a)
b) Angle ECF is an exterior angle of triangle ABC, and so, its measure is the sum of the measures of angles A and B. Let the circles through ECF and BDE meet at point M. Then angle FME and angle ECF are supplementary. Also, the angles EMD and EBD (i.e. angle B) are congruent because they inscribe the same arc in circle BDE. Now, the measure of angle FME = \pi - measure of angle ECF = \pi - measure of angle A - measure of angle B. So, we have, \pi - m(A) = m(FME) + m(B) = m(FME) + m(EMD) = m(FMD). Thus, angles FMD and A are supplementary, implying that M is on the circle determined by A, F and D.
17. Show that, on the Euler line, the centroid and the orthocenter divide internally and externally in the same ratio the segment whose endpoints are the circumcenter and the center of the nine-points circle.
Solution: Since the centroid, G, trisects HO and the center of the nine-points circle N bisects HO, we have the following: If the length of NG is denoted by x, then GO is 2x and HN is 3x. The internal ratio NG/GO = x/2x = 1/2, and the external ratio NH/HO = 3x/6x = 1/2. Thus, G and H divide the segment NO internally and externally in the same ratio.
19. Prove that if the Miquel point is on the circumcircle, then the three points on the sides of the triangle are collinear.
Solution:Refer to the diagram of exercise 15a. Since, M is on the circumcircle, angles CMB and A are supplementary. Since A,D, M and F are on the same circle, angles FMD and A are supplementary. Therefore, angles CMB and FMD are congruent. Since these two angles overlap in angle CMD, we have that angles CMF and DMB are congruent. Now angle CEF is congruent to CMF because they inscribe the same arc in circle CEF. Also, angle DEB and angle DMB are congruent because they inscribe the same arc in circle DEB. Thus, CEF and DEB are congruent. As, CEB is a line, DEF must also lie on a line (opposite angles are equal).