## Wednesday, October 22, 2014

### Geometry Problem 1051: Triangle, Square, Semicircle, Arc, Area

Level: Mathematics Education, High School, Honors Geometry, College.

Problem submitted by Kadir Latintas, Math teacher in Emirdag, Turkey

Click the figure below to enlarge the figure. 1. Assume square ABCD defines a square grid with A as origin and D = (1,0). Add two more grids: a blue grid such that AD = blue(2,-1) and a brown one where AD = brown(4,1). See http://bleaug.free.fr/gogeometry/1051.png

In this notation E=brown(0,2) and F=blue(1,1). With grid calculation, we see that EF slope = brown(1,3) = blue(5,3).

Let AH1=brown(1,3) // EF therefore area(EFH1)=S1. Since EH1=brown(1,1) and EC=brown(3,3) thus S3=3*S1

Let FB'=blue(0,-1) therefore area(EFB')=S2. H2 on EC with B'H2//EF implies EH2=brown(2.5, 2.5) thus S2=2.5*S1

=> S1+S2+S3=6.5*S1

Take E'F' symmetric of EF relative square centre. See http://bleaug.free.fr/gogeometry/1051b.png
Since S1, S2, S3 share the same base, sum of their areas = base*sum of heights. From picture, we see that S1+S2+S3=S4

tan(ADE/2)=1/4 and tan(CDF/2)=1/2. By composing tangents we get tan(FDE)=13/84 => sin(FDE)=13/85

S4=S/2*sin(FDE)=13/170*S

S=85*S1=34*S2

bleaug

2. Assume A:(0,0), B:(0,4), C:(4,4) & D:(4,0) i.e a square of side 4 units.
So E is the intersection of:(x-4)² + y²=16 & x²+(y-1)²=1 and thus E is (8/17.32/17).
Similarly, F is the intersection of: (x-4)² + y²= 16 and (x-2)²+(y-4)²=4 and thus F is (4/5.12/5)
Triangle Area: A=(1/2)[x1(y2- y3)+x2(y3- y1)+x3(y1- y2)]
(x1,y1)n can taken as the origin (0,0) while all other necessary points are known. Thus we can calculate S1 = 16/85, S2=40/85 & S3 = 48/85 while S4 = 104/85
It is easy to see that: S4 = S1 + S2 + S3 and that S = 85 * S1 = 34* S2.