Triangle Area

On page 464 of Geometry, there is a challenge problem. Three segments through a point P and parallel to the sides of a triangle divide the whole triangle into six small regions. The three small triangular regions have the areas 9, 16, and 25. Find the area of the whole triangle.

(1) Prove the small triangles are similar to the whole triangle.
(2) Use the area ratio and the scale factor equation
(3) Use the segment additional property.
(3) Solve for the area.

A = { 9^(1/2) + 16^(1/2) +25^(1/2) } ^2

ChrisRao/饶旭东

Ptolemy Theorem

Ptolemy Theorem states that in an inscribed quadrilateral, the sum of the products of its opposite sides is equal to the product of its diagonals, or ac + bd = ef

(1) For each side(a, b, c, and d) and its corresponding inscribed angle (A, B, C, and D) of the inscribed qudrilateral, we have :

a = 2R sin A
b = 2R sin B
c = 2R sin C

For each diagonal (e and f) and its corresponding inscribed angle (E and F) of the inscribed qudrilateral, we have:

e = 2R sin E, E = A + B
f = 2R sin F, F = A + D

(2) In order to prove ac + bd = ef, we have to prove:

sin A sin C + sin B sin D = sin (A+B) sin (A+D)

sin A sin C = 0.5 [ cos (A-C) - cos (A+C) ]
sin B sin D = 0.5 [ cos (B-D) - cos (B+D) ]

(3) If we notice that

(A+C) + (B+D) = 180
cos (A+C) = - cos (B+D)

then we have:

sin A sin C + sin B sin D = 0.5 [ cos (A-C) + cos (B-D) ]

(4) If we also notice that

A + B + A + D=A +(180-C) = 180 + (A-C)
cos (180+A-C) = -cos(A-C)

sin (A+B) sin (A+D) = 0.5 [ cos (B-D) + cos (A-C) ]

(5) So, sin A sin C + sin B sin D = sin (A+B) sin (A+D)

(6) Therefore, ad + cd = ef

CHRISRAO/饶旭东

怀广济游长江上海西安

第一次对对联有兴趣是因为我少时好友教我对对联。她教我的第一个对联就是：山石岩前古木枯此木为柴，白水泉边女子好少女更妙。从此在我的映像里她就成了那泉边的少女。从那以后学了很多对联。学了对联以后才知道绝对的来历。原来绝对就是没有对出联来的一半对联。

到现在还记得的有两幅绝对。其中之一就是：望江楼，望江流，望江楼上望江流，江流千古，江楼千古。这个绝对有名，所以在网上已经有很多对出来的下对。很多人觉得这个绝对已经解决了。但是我对百度出来的下对还没有觉得满意的。在香港有一条河流叫双鱼河，可以据此作下对：双鱼河，双鱼游，双鱼河下双鱼游，鱼游几里，鱼河几里。把这两联摆在一起，显得更加工整。

望江楼，望江流，望江楼上望江流，江流千古，江楼千古。
双鱼河，双鱼游，双鱼河下双鱼游，鱼游几里，鱼河几里。

还有一个绝对，就是家乡广济县人知道的：出宿松泛太湖望江东流。这么多年来我一直没有找到相关的下对。经过几十年的漂洋过海背井离乡的生活，到今天才觉得自己有了点心得：怀广济游长江上海西安。把这两句摆在一起就更好看一点儿。

出宿松泛太湖望江东流，
怀广济游长江上海西安。

Construct an Equilateral Triangle

How to construct an Equilateral Triangle?

42. Construct an Equilateral Triangle whose area is equal to the area of a scalene triangle ABC. (Page 433 of Geometry).

There are three steps to solve this problem. Suppose the scalene triangle is characterized with a base b and a height h. The first step involves a semicircle of the diameter (b+h) with an inscribed right triangle. The last two steps involve the geometric means of right triangles.

(1) As we know from the Corollary 2 of Theorem 9-7 that an angle inscribed in a semicircle is a right angle. The two sides of the right angle and the diameter form a right triangle. Let the diameter of the semicircle equals to the length of b+h of the given scalene triangle.

(2) We know from the corollary 1 of Theorem 8-1 that if the altitude is drawn to the hypotenuse of a right triangle, then the length of the altitude is the geometric means between the two segments of the hypotenuse. That is to say x = sqrt (b*h).

(3) We know from the corollary 2 of Theorem 8-1 that if the altitude is drawn to the hypotenuse of a right triangle, then each leg is the geometric means between the hypotenuse and the segment of the hypotenuse that is adjacent to that leg. That is to say x = sqrt (k*s), where k and s are the altitude and base of the equilateral triangle we need to construct. This right triangle should be a special 30-60-90 right triangle. The x is the length of one leg of the special right triangle.

Since we know the length of each side of the equilateral triangle, it should be easy to construct the equilateral triangle.

CHRISRAO/饶旭东