1. Use the notation of Figure 4.3 to name all of the triangles whose vertices are three of the given points and whose orthocenter is the fourth distinct point.
Solution: ABC orthocenter H, ABH orthocenter C, BCH orthocenter A, ACH orthocenter B.
2. Could the centroid be outside the triangle?
Ans: No Solution:The intersection of any two medians is inside the triangle. Consider any median. The two vertices not on this median are on opposite sides of the median, so, one of these vertices must be on the opposite side of median to the side of the triangle opposite this vertex. When the median from this vertex is drawn, it must intersect the first median before it intersects the midpoint of the opposite side, so the point of intersection is inside the triangle.
3. Could the incenter be outside the triangle?
Ans:No.Solution: You can reason in a manner analogous to problem 2.
4. Could the circumcenter be outside the triangle?
Ans:Yes. Solution: This happens in any triangle containing an obtuse angle.
5.Could the incircle and circumcircle have points in common?
Ans: No. Solution: The incircle is contained in the triangle while the circumcircle only meets the triangle at the vertices. If they had a point in common, it would have to be a vertex of the triangle. But the incircle does not contain any of the vertices of the triangle.
6. Where is the orthocenter of a right triangle?
Ans:At the vertex which is the right angle. Solution: The sides of a right triangle are also the altitudes drawn from the two vertices on the hypotenuse. They meet at the right angle, so it must be the orthocenter.
7. Under what conditions would the orthocenter of a triangle lie outside the triangle?
Solution: Any obtuse triangle. The altitudes drawn to the sides of the obtuse angle always lie outside the triangle.
8. How many excenters does a triangle have?
Ans: 3. Solution: Excenters are determined by pairs of external bisectors. There are 3 such pairs, giving 3 excenters.
10. Find the area of the trianglular region if the sides of the triangle are a = 7, b = 12 and c = 15.
Ans:~ 41.231 ... Solution: The semi-perimeter is (1/2)(7 + 12 + 15) = 17. By Heron's formula, the area is sqrt( 17(17-7)(17-12)(17-15)) = sqrt(17(10)(5)(2)) = sqrt(1700) = 41.231...
23. Prove that the internal and external angle bisectors at a vertex of a triangle are perpendicular.
Solution:
31. A paradox that results from an incorrect assumption in Figure 4.10 is the proof that any triangle is isosceles. Let ABC be any triangle, with D the intersection of the angle bisector at B and the perpendicular bisector of AC. DE and DF are perpendicular to the sides.
Solution: