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.

 Problem 371: 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?

  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


    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
    Sri Lanka

  6. 2nd Pure Geometry Solution, much easier

    < NGD = < NFG = < MCG since FMGC is concyclic as a result of congruent triangles AGD & FCD

    So CM // GN & N is the mid point of DM

    So S1 = S(MGD)/2 = S(FCD)/10 since Tr.s MGD & FCD are similar & FD^2 = 5.GD^2

    But S(FCD) = S/4 and the result follows

    Sumith Peiris
    Sri Lanka

  7. For ease of calculations, let the length of the square be 2a units
    Since ADG congruent to DCF & AD perpendicular to DC => AG perpendicular to FD => NMG is a right triangle --------(1)
    Let m(MDG)=α and connect EG,FH and EN
    similarly m(AGE)=α and m(HFD)=α (since AGE,HFD,DAG & CDF are congruent triangles)
    since m(HEG)=45 and m(HEN)=m(HFN)=m(HFD)=α => m(NEG)=45-α ---------(2)
    m(ENG)=90 (angle in semi-circle) ---------(3)
    From (2) & (3) m(EGN)=45+α
    => m(MGN) = 45 (as m(AGE)=α)
    Hence (1) is an isosceles right triangle and hence MG=MN ----------(4)

    Denote I as point of intersection of EN and AG & observe that triangles GEI and DNG are similar
    Since EG=2.DG
    => GI=2.DN
    Since m(ENF)=m(INM)=45 and NM perpendicular to IG=> M is midpoint of IG and MG=MN=ND
    Apply pythogrous to MGD => MG=MN=a/sqrt(5)
    Area of MGN=S1=a^2/10= 4a^2/40=S/40
    =>S=40.S1 Q.E.D

  8. Join O to G. FD meet OG at P. ∆AMD ~ ∆MGD => ˂ M = 90°
    From similarity if MG = x => MP = x/2, MD = 2x
    => S(PGD) = 1/2(xˑ5x/2) = 1/16 S
    => 1/2 (x²) = 1/40 S

  9. See graph here.
    ∠CDF = ∠DAG ⇒ ∆DMG and ∆ADG are similar ⇒ ∆GMN is rectangle in M
    ∆ENG rectangle in N and ∠ENF = 45° ⇒ ∆GMN is isoscele in M and MN = MG = DG/sqrt(5)
    Hence S1 = 1/2.MG.MN = DG∧2/10 = S/40 QED