Saturday, September 1, 2012

Problem 801: Equilateral Triangles, Cevian, Midpoint, Distance, Measurement, Metric Relations

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

Click the figure below to see the complete problem 801.

Online Geometry Problem 801: Equilateral Triangles, Cevian, Midpoint, Distance, Measurement, Metric Relations.

Posted by Antonio Gutierrez at 10:55 AM
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10 comments:

  1. AnonymousSeptember 2, 2012 at 7:29 AM

    Let AB=a,CD=b and BD=y
    1- Δ BCE
    a²+b²+ab=100
    2- Δ BCD
    a²+b²-ab=y²
    2(a²+b²)=100+y²
    3- ΔBDC => theorem of apollonius
    2(a²+b²)=4x²+y²
    4x²+y²=100+y²
    4x²=100
    X=5

    Erina-NJ

    ReplyDelete
  2. AnonymousSeptember 2, 2012 at 10:35 AM

    ED meets AB at H. CH=BD and CF=BG, where F and G are mindpoints of BD and CH respectivly. It is obvious that CF=BG=5.
    You can see this here: http://mtz256.files.wordpress.com/2012/09/801.jpg

    ReplyDelete
  3. Peter TranSeptember 2, 2012 at 3:56 PM

    Let AB=BC=Ac=a and CD=CE=DE=b
    Apply cosine formula in triangle BCE we have BE^2=a^2+b^2+a.b= 10^2
    Apply cosine formula in triangle BCD we have BD^2=a^2+b^2-a.b
    In triangle BCD apply formula for calculate median we have 2.CF^2=BC^2+CD^2-BD^2/2
    Or 2.x^2=1/2(a^2+b^2+a.b)= ½.BE^2=50
    So x=5

    ReplyDelete
  4. AnonymousSeptember 3, 2012 at 7:00 AM

    1) continue CF until CF=FP => CF = CP/2
    BCDP is a paralellogram
    2) E, D, P are collinear. BCEP is an isosceles trapezoid.
    PC=BE=10 => CF=5


    Erina-NJ

    ReplyDelete
  5. Antonio GutierrezSeptember 3, 2012 at 2:07 PM

    Problem 801 solution: This solution was submitted by Michael Tsourakakis from Greece

    Thanks Mixalis.

    ReplyDelete
  6. AnonymousSeptember 3, 2012 at 2:36 PM

    Sea M el punto medio del lado BC. Entonces el segmento FM es paralelo a DC y mide DC/2, y el triángulo CFM es semejante al triángulo BEC, por el criterio LAL, CM=BC/2, FM=DC/2=EC/2, ángulo FMC = ángulo ECB = 120º, así pues x=CF=BE/2=5.

    Migue.

    ReplyDelete
  7. PravinSeptember 5, 2012 at 1:49 AM

    Complete the parallelogram BCDG.
    Triangles BCG, BCE are congruent.
    2x = GC = BE = 10.
    x = 5

    ReplyDelete
  8. Prof Radu Ion,Sc.Gim.Bozioru,BuzauOctober 29, 2012 at 12:01 PM

    Construind simetricul punctului D fata de C si notandu-l cu P obtinem ca
    Δ BCE=Δ BCP (LUL) BE=BP si CF linie mijlocie in
    Δ BDP <=> x=CF=BE/2=5.

    ReplyDelete
  9. Ivan BazarovFebruary 15, 2014 at 6:25 PM

    BP is parallel to FC such that P is on AC. DC=CP but also DC=EC so CP=CE. <BCE=60+60=120, but also <BCP=180=60=120 so <BCE=BCP. These equalities show that BCE and BCP are congruent, making BC=10=BP=2x so x=5.

    ReplyDelete
  10. Sumith PeirisFebruary 4, 2016 at 10:01 PM

    Simple pure geometry solution

    Draw FG // to DC where G is on BC

    Consider Tr.s BCE & FGC

    < C = < G = 120
    BC = 2 GC
    CE = 2 FG ( last 2 results from applying mid point theorem to Tr. BCD)

    The 2 triangles are therefore obviously similar and the sides of Tr. FGC are half that of Tr. BCE

    Hence x = 10/2 = 5

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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