Non-right triangles and trigonometry #4 - Questions and Answers

Question 1 What is the inradius of the octahedron with side length 𝑎 a? Determine the height of one of two constituent square pyramids by considering a right triangle, using the fact that the height of the equilateral triangle is 𝑎 3 2 a 2 3 ​ ​ . (Incidentally, by this, you obtain the circumradius of the Octahedron.) Now cut the (solid) octahedron along 4 of these latter heights, i.e., across two non-adjacent vertices and two midpoints of parallel sides of the "square base" (Figure 1). Consider the resultant rhombic polygon, whose side length is 𝑎 3 2 a 2 3 ​ ​ . Choose a convenient pair of adjacent corners 𝐴 A and 𝐵 B, s.t. the angle in 𝐴 A is obtuse. Then the insphere touches 𝐴 𝐵 AB, say in 𝑀 M. So the question becomes: Is there a still simpler proof? Answer 1 Step 1 An analytical response: Finding the sphere centered at the origin that is tangent to every face of the octahedron is the task at hand if we center the octahedron at the origin. One face is all that is required to understand symmetry. A model with an equation: 4 𝑥 + 𝐵 𝑦 + 𝐶 𝑧 + 𝐷 = 0 4x+By+Cz+D=0 is tangent to the sphere: 𝑥 2 + 𝑦 2 + 𝑧 2 = 𝑟 2 x 2 +y 2 +z 2 =r 2 if the coefficients satisfy the dual quadric equation: 𝐴 2 + 𝐵 2 + 𝐶 2 = 𝑟 2 A 2 +B 2 +C 2 =r 2 W.L.O.G. we can place the three vertices at: ( 𝑎 2 , 𝑎 2 , 0 ) , ( − 𝑎 2 , 𝑎 2 , 0 ) , and  ( 0 , 0 , 𝑎 2 ) ( 2 a ​ , 2 a ​ ,0),(− 2 a ​ , 2 a ​ ,0),and (0,0, 2 ​ a ​ ) The plane through these points has the equation: ∣ 𝑥 𝑦 𝑧 1 𝑎 2 𝑎 2 0 1 − 𝑎 2 𝑎 2 0 1 0 0 𝑎 2 1 ∣ = 0 ​ x 2 a ​ − 2 a ​ 0 ​ y 2 a ​ 2 a ​ 0 ​ z 0 0 2 ​ a ​ ​ 1 1 1 1 ​ ​ =0 Which simplifies to 2 𝑦 + 2 𝑧 = 𝑎 2y+ 2 ​ z=a. Substituting the coefficients of this plane equation into: 𝐴 2 + 𝐵 2 + 𝐶 2 = 𝑟 2 A 2 +B 2 +C 2 =r 2 produces: 𝑎 2 3 = 6    ⟹    𝑟 = 𝑎 6 3 a 2 ​ =6⟹r= 6 ​ a ​ Answer 2 Step 1 Consider this full elementary geometrical derivation: Inscribe into the octahedron its dual cube. Then the searched-for inradius of the octahedron would be the circumradius of that cube. Let the edge length of the octahedron be 𝑎 a, and that of the inscribed dual cube be 𝑏 b. Next, consider a variation of that dual cube as an accordingly oriented square prism of varying height, still with vertices on the faces of the octahedron. That prism would have base sides of size: 𝑏 = 𝑎 2 ⋅ 𝑥 b= 2 ​ a ​ ⋅x Where 𝑡 t is some variation parameter, starting with 𝑡 = 0 t=0 at the top and ending with 𝑡 = 1 t=1 at the equatorial section of the octahedron. The corresponding height of that prism then is: ℎ = 2 𝑎 ( 1 − 𝑡 ) h= 2a ​ (1−t) For the prism to become a cube, we have: 𝑏 = ℎ or 1 2 𝑎 𝑥 = 2 𝑎 ( 1 − 𝑡 ) b=hor 2 ​ 1 ​ ax= 2a ​ (1−t) Solving for 𝑡 t results in: 𝑡 = 1 2 , 𝑏 = 𝑎 3 t= 2 1 ​ ,b= 3 ​ a ​ Now the inradius of the 𝑎 a-sized octahedron is the circumradius of the 𝑐 c-sized cube, and thus clearly: 𝑟 = 3 2 𝑐 = 3 𝑎 2 − 4 𝑎 6 𝑎 r= 2 3 ​ ​ c= 6 3a 2 −4a ​ ​ a Answer 3 Sketch of idea: I tend to think of it as follows. Consider an octahedron with points on the 𝑥 , 𝑦 , 𝑧 x,y,z axes, at coordinates ± 1 ±1. Then by symmetry, the inradius is the distance from the origin to the point ( 1 3 , 1 3 , 1 3 ) ( 3 1 ​ , 3 1 ​ , 3 1 ​ ), which is 𝑎 3 3 a ​ . By similarity, the inradius of an octahedron with side length 𝑎 a is: 𝑟 = 𝑎 3 6 r= 3 6 ​ a ​ Question 2 One of the largest issues in ancient mathematics was accuracy—nobody had calculators that went out ten decimal places, and accuracy generally got worse as the numbers got larger. The famous Eratosthenes experiment. Given two similar triangles, one with small measurements that can be accurately determined, and the other with large measurements, but at least one is known with accuracy, can the other two measurements be measured? Explain and give an example. The similarity of triangles gives rise to trigonometry. Answer: Calculations using recurring or non-terminating decimals provide an accuracy issue. Trigonometry enables us to describe values in ratio form that may then be utilized for new calculations that provide more precise answers by using ratios like sin, cos, or tan. We have Pythagorean theorem 𝑐 2 = 𝑎 2 + 𝑏 2 c 2 =a 2 +b 2 . Trigonometry can therefore be used in a wide range of applications to accurately represent larger repeating and non-terminating decimals. Question 3 State the form of the law of cosines used to solve a triangle for which all three sides are given (SSS). Answer: There are basically three laws of the cosine which are given as follows: 𝑎 2 = 𝑏 2 + 𝑐 2 − 2 𝑏 𝑐 cos ⁡ 𝐴 a 2 =b 2 +c 2 −2bccosA 𝑏 2 = 𝑎 2 + 𝑐 2 − 2 𝑎 𝑐 cos ⁡ 𝐵 b 2 =a 2 +c 2 −2accosB 𝑐 2 = 𝑎 2 + 𝑏 2 − 2 𝑎 𝑏 cos ⁡ 𝐶 c 2 =a 2 +b 2 −2abcosC Refer to the figure for the indication of the angles and sides.