Reflect the triangle at the top of the pentagram over its base, forming, as shown in the figure. Now, . Then the darker shaded trapezoid has the same area as three of the outside triangles and . It can also be shown that , so the area of the lightly shaded portion that is in the pentagram but not in the darker region is also the same as three of the outside triangles and . Thus, the shaded ½ region is the total area. Let denote the circumcenter of the triangle which lies inside the square. We first note A that in an equilateral triangle of side length 1, its altitudes goes through the circumcenter and the circumcenter divides the altitude into two segments of lengths and . Thus, is convex and its area is the sum of the area of triangle and the area of ABCD ABD triangle The length of is . In triangle the altitude corresponding to base CBD. BD ABD, is . Thus, the area of is . Next, in triangle the altitude corresponding to BD ABD CBD, base is Thus, the area of is . Therefore, the area of is . (I used a kite and BD CBD ABCD thought it easier.) Since is the incenter, we know that is the angle I BE bisector of By the angle bisector theorem, . ∠ABC. Also, we know that = 20 Thus, . Because CE AE. − D is the point of tangency, we also know that = AD s where s is the semiperimeter, which is 30. BC, − This implies that = 9. Finally, = = AD DE AE AD − ½ Let the triangle be denoted with = 45◦, = 60◦, and = 75◦. We first find ABC, ∠A ∠B ∠C side Drop an altitude onto side and note that is a 45-45-90 triangle and AB. D AB ACD is a 30-60-90 triangle. Therefore, and. Thus,Since is an altitude, the BCD AD = CD CD area of is. Since the area of is, it follows that and hence = 2. Now, . Since ABC ABC AB is a 45-45-90 triangle,. Since is a 30-60-90 triangle, . Thus, the perimeter of ACD BCD is . ABC
Reflect the triangle at the top of the pentagram over its base, forming, as shown in the figure. Now, . Then the darker shaded trapezoid has the same area as three of the outside triangles and . It can also be shown that , so the area of the lightly shaded portion that is in the pentagram but not in the darker region is also the same as three of the outside triangles and . Thus, the shaded region is ½ the total area. Let A denote the circumcenter of the triangle which lies inside the square. We first note that in an equilateral triangle of side length 1, its altitudes goes through the circumcenter and the circumcenter divides the altitude into two segments of lengths and . Thus, ABCD is convex and its area is the sum of the area of triangle ABD and the area of triangle CBD. The length of BD is . In triangle ABD, the altitude corresponding to base BD is . Thus, the area of ABD is . Next, in triangle CBD, the altitude corresponding to base BD is Thus, the area of CBD is . Therefore, the area of ABCD is . (I used a kite and thought it easier.) Since I is the incenter, we know that BE is the angle bisector of ∠ABC. By the angle bisector theorem, . Also, we know that CE = 20 − AE. Thus, . Because D is the point of tangency, we also know that AD = s − BC, where s is the semiperimeter, which is 30. This implies that AD = 9. Finally, DE = AE − AD = ½ Let the triangle be denoted ABC, with ∠A = 45◦, ∠B = 60◦, and ∠C = 75◦. We first find side AB. Drop an altitude D onto side AB and note that ACD is a 45-45-90 triangle and BCD is a 30-60-90 triangle. Therefore, AD = CD and. Thus,Since CD is an altitude, the area of ABC is. Since the area of ABC is, it follows that and hence AB = 2. Now, . Since ACD is a 45-45-90 triangle,. Since BCD is a 30-60-90 triangle, . Thus, the perimeter of ABC is .
ARML_Practice_9-14-14.docx