Tuesday, May 27, 2008

Geometry Problem 110

See complete Problem 110
Contact Triangle, Area, Incircle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.


  1. For trigonometry lovers!
    Triangles ADE,BDF,EFC are isoscele,hence we know the angles.
    after substitution
    after substitution and simplification
    b=2RsinB and 2sin(B/2)cos(B/2)=sinB
    with Heron's formula we can say
    the transformation of (p-b) is made by

  2. Let A=Area of Tr. ABC. Using standard notation, A^2=s(s-a)(s-b)(s-c). Also A=r*s where r=inradius and s=semiperimeter as usual. Thus,
    (rs)^2=s(s-a)(s-b)(s-c) or r^2*s=(s-a)(s-b)(s-c). We now use the result of Problem #82 whereby S/A=r/2R or S=rA/2R=(r^2*s)/2R =(s-a)(s-b)(s-c)/2R or 2/S=4R/[(s-a)(s-b)(s-c)]-----(1).
    Now draw a perpendicular,h, from D to BC. We've S1=h*AD/2=h(s-a)/2. Like wise, S2=h(s-c)/2 since Tr. BDF is isosceles and thus 1/S1+1/S2=2[1/(s-a)+1/(s-c)]/h=2b/[sin(B)(s-a)(s-b)(s-c)]. But b/sin(B)=2R hence 2/S =4R/[(s-a)(s-b)(s-c)]-----(2). By equations (1) & (2), we've 2/S = 1/S1 + 1/S2 ; hence etc.
    Antonio, if there's still an easier way to do this, would you be kind enough to show us the same?

  3. Suggestions for another solution:

    1. The area of a triangle is equal to one half of the product of its altitude and base.

    2. Similar triangles.

    3. The two tangents to a circle from an external point are congruent.

  4. One more solution as suggested by Antonio: In the given diagram, let FD and CA (both extended) meet in V and let angle DVA=Θ & VA=x. Draw perpendiculars h1, h & h2 to FD extended from A, E & C resply. Now proving that S=2S1/(S1+S2) is tantamount to proving h = 2h1h2/(h1+h2) since the three triangles have a common base FD.
    Apply Menelaus's Theorem to Tr. ABC with FDV as the transversal, we've:x/(x+b)*(s-c)/(s-b)*(s-b)/(s-a)=1(ignoring the - sign which is unimportant here). This gives us x/(x+b) =(s-a)/(s-c).
    Note that AD=AE=(s-a), CE=CF=(s-c) & BF=BD=(s-b) where 2s=a+b+c. Now, h1=(x)sin(Θ), h2=(x+b)sin(Θ) while h=(x+s-a)sin(Θ). Substitute the values above to eliminate x & s to obtain: h=[b^2 -(a-c)^2]sin(Θ)/2(a-c) & 2h1h2/(h1+h2) also =[b^2-(a-c)^2]sin(Θ)/2(a-c). In other words, h = 2h1h2/(h1+h2) or S = 2S1S2/(S1+S2)

  5. I propose an alternative solution to problem 110, following the idea showed by Ajit, with slightly different termination.

    Let h, h1 and h2 be the altitudes EG, AJ and CK, of triangles DEF, ADF and DFC, respectively, related to the common base DF. Take H on the extension of AJ, with AH = 2*h1. Take I such that HI is parallel to DF and CI is perpendicular to DF. We have CII = h1 + h2.
    If we prove that h = 2*h1*h2/(h1 + h2), this implies that S = 2*S1*S2/(S1 + S2).

    The extension of KJ meets the extension of CA at V. The extension of IH meets the extension of CA at W. Putting AV = x, we have AW = 2x. Let’s use the Menelaus’ theorem to triangle ABC, with the transversal DF, which meets AB at D, BC at F, and CA at V: (AD/DB)(BF/FC)(CV/VA) = 1, or [(s-a)/(s-b)]*[(s-b)/(s-c)]*[(x+b)/x] = 1, where s is the semi perimeter of ABC. Thus, x/(x+b) = (s-a)/(s-c), and x = b*(s-a)/(a-c) (1).

    We have EG/CK = h/h2 = VE/VC = (x+s-a)/(x+b). Substituting (1) and simplifying the result, we get h/h2 = 2*(s-a)/b.

    We have also AH/CI = 2*h1/(h1+h2) = AW/CW = 2x/(2x + b). Substituting (1) and simplifying the result, we get 2*h1/(h1+h2) = 2*(s-a)/b.
    So, h/h2 = 2*h1/(h1+h2) and finally h = 2*h1*h2/(h1 + h2).