Monday, June 16, 2014

Geometry Problem 1024: Contact, Gergonne Triangle, Point on an arc of Incircle, Perpendicular, Distances

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1024.

Online Math: Geometry Problem 1024: Contact, Gergonne Triangle, Point on an arc of Incircle, Perpendicular, Distances

3 comments:

  1. Denote inscribed angles of arcs PTa, PTc, and TcTb to be a, b, and c respectively.
    a1=PTa*sin(a)=(PTa)^2/(2R)
    b1=PTb*sin(b+c)=(PTb)^2/(2R)
    c1=PTc*sin(b)=(PTc)^2/(2R)
    Plugging these into equation to be proved yields Ptolemy's Theorem in quadrilateral PTaTbTc.

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  2. Can someone post a solution to this question, I'm having a difficult time proving it

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  3. https://goo.gl/photos/8vk1AQ8RWkqjdc3Q9

    Let PQ is the diameter of circle O and R= radius of circle O
    Define a as ∠ (PTaB)= ∠ (PQTa)
    . b=∠ (BTcP)= ∠ (PQTc)
    . c= ∠ (ATbTc)= ∠ (TbQTc)
    See sketch for details
    We have
    Sin(a)=PTa/2.R
    Sin(b)=PTc/2R
    Sin(b+c)= PTb/2R
    a1=PTa*sin(a)=(PTa)^2/(2R) => PTa= sqrt(2.R.a1)
    b1=PTb*sin(b+c)=(PTb)^2/(2R)=> PTb= sqrt(2.r.b1)
    c1=PTc*sin(b)=(PTc)^2/(2R) => PTc= sqrt(2.R.c1)
    Apply Ptolemy’s theorem in quad. PTaTbTc
    we have TaTc. PTb= PTc.TaTb+ PTa.TbTc
    Plug these values in above equation we will get the result

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