## Thursday, October 22, 2009

### Geometry Problem 371: Square, Inscribed circle, Triangle, Area

Proposed Problem
Click the figure below to see the complete problem 371 about Square, Inscribed circle, Triangle, Area.

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Complete Geometry Problem 371
Level: High School, SAT Prep, College geometry

1. Let O be (0,0) and G be (a,0). This makes F:(0,a), A:(-a,-a) and G:(a,-a). Determine equations of FD and AG as 2x + y = a & x - 2y = a. Their intersection is M: (3a/5, -a/5). The circle at O is x^2 + y^2 = a ^2 and its intersection with FD gives us N as (4a/5,-3a/5).
Area of a triangle with vertices at (x1,y1), (x2,y2) and (x3,y3) is given by: (1/2)[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]. Plugging in the values, we get area of tr. GMN as (1/2)[a(-a/5+3a/5)+(3a/5)(-3a/5)+(4a/5)(a/5)]= a^2/10 = 4a^2/40.
In other wods, S1 = S/40
Antonio, how may one solve this without using analytical geometry?
Ajit: ajitathle@gmail.com

2. Some ideas:
Right triangles
Congruent triangles
Similar triangles
Angles in a circle
Area of a triangle and square

3. How about thia for a plane geometry solution?
Let the square side=2a. Tr. GAD & Tr. FDC are congruent and hence angle GAD = ang. FDC. But ang. FDC + ang. ADF = 90 and hence ang. GAD + ang. ADF = 90 and, therefore, ang. AMD = 90. Now AG=V(4a^2+a^2)=V5a=FD where V = square root. Further, AD*DG/AG =MD which gives us MD=2a/V5. Moreover, ND*DF=DG^2 or ND =a^2/(V5a)=a/V5 or N bisects MD which makes Tr.GMN=Tr.GMD/2 -----(1).
Now Tr. GMD and Tr. AFC are similar and hence Tr.GMD/a^2=(2a/V5)^2/4a^2 or Tr. GMD =4a^2/20 which makes Tr. GMN= 4a^2/40 by (1) above or S1=S/40, as required
Ajit: ajitathle@gmail.com

Solution without analitycal geometry. Fales theorem - main idea.

5. Pure Geometry solution

Let AB = a and MG = p

Tr.s FCD and AGD are congruent SAS

Hence < GDM = < GAD and so < GMD = 90.

So p = AG/GD^2 = a /(2sqrt5)

< CGF = 45 hence < MNG = 45

So S1 = p^2/2 = a^2/(2X(2sqrt)^2) = a^2/40

Hence S = 40S1

Sumith Peiris
Moratuwa
Sri Lanka