Monday, March 24, 2014

Geometry Problem 997. Equilateral Triangle, Vertices, Three Parallel Lines, Distance, Metric Relations

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 997.

Online Geometry Problem 997. Equilateral Triangle, Vertices, Three Parallel Lines, Distance, Metric Relations

Posted by Antonio Gutierrez at 3:17 PM

8 comments:

  1. Jacob HA (EMK2000)March 24, 2014 at 9:56 PM

    Let z(P) be the complex number representing P.
    Let z(A)=0, z(C)=a+15i where a²+15²=x².

    Then z(B)=(a+15i)(1/2+√3/2 i)=(a/2−15√3/2)+(15/2+√3/2 a)i

    Thus 15/2+√3/2 a=9, so a=√3.
    x²=a²+15²=228, x=√228=2√57.

    ReplyDelete
  2. HenkieMarch 25, 2014 at 4:04 PM

    Call α = angle(DCB), then angle(EBA) = 30+α.
    sin α = 6/BC = 6/AB and cos (30 + α) = 9/AB
    Using the formula for cos(x+y) gives AB = 18/(√3cosα - sinα) = 18/(√3cosα - 6/AB)
    This gives cosα = 24/(√3*AB), or cos²α = 192/AB².
    Using cos²α + sin²α = 1 gives 36/AB² + 192/AB² = 1
    Or: AB² = 228 → AB = √228

    ReplyDelete
  3. MaxDecember 22, 2014 at 2:47 PM

    Define E as the origin of a graph, so that B is at (0, 9) and A is at (a, 0). Then C is at the intersection of these two circles:

    (81+a²) = (x-a)² + y² (circle with radius length AB centered at A)
    and
    (81+a²) = x² + (y-9)² (circle with radius length AB centered at B)

    Rewriting, we get y = ±√(81 - x² + 2ax) and y = 9 ± √(a² + 81 - x²)

    Furthermore, the y-value of C is at 15.

    Solving both expressions above for y = 15 will give us all of the x-values, in terms of a, that result in the circles intersecting with the line y = 15. Setting the x-values equal to each other will tell us the value of a at which the circles intersect on y = 15.

    Inspecting the graph makes it clear that the functions intersect in the upper portions of both circles, so we need to focus on:

    y = √(81 - x² + 2ax) and y = 9 + √(a² + 81 - x²)

    15 = √(81 - x² + 2ax) can be simplified to 144 = 2ax - x².

    y = 9 + √(a² + 81 - x²) can be simplified to x = √(a² + 45)

    Substituting gives us 144 = 2a√(a² + 45) - (√(a² + 45))²

    After lots of simplifying, we get that 0 = a^4 - 66a² - 11907
    That factors to 0 = (a² + 81)(a² - 147)
    Which factors to 0 = (a + 9i)(a - 9i)(a + 7√3)(a - 7√3)

    So the solution in the first quadrant is a = 7√3, in other words that the coordinates of A are (0, 7√3)

    The Pythagorean theorem gives the length of AB: 9² + (7√3)² = AB²; AB = √288 = 2√57

    ReplyDelete
  4. mpakiMarch 7, 2015 at 4:44 AM

    Let F be the reflection of B with respect to the line CD. Then the circle (C,x) passes through the points A,B and F.
    So angle(AFB) = 60/2 = 30
    From the right triangle FEA we get EA = FE*tan(30) = 7√3
    From the right triangle BEA we get x^2 = BE^2 + EA^2 = 81 + (7√3)^2 = 228

    ReplyDelete
  5. c.t.e.oFebruary 20, 2019 at 10:37 AM

    Draw altitude BH, draw MH // EA (M midpoint), draw HT ꓕ EA
    => √BMH ~ ∆ATH => AT = √3/2, from ∆ATH => x²= 228

    ReplyDelete
  6. MarcoApril 13, 2024 at 10:30 PM

    Let <DCB=a
    sina=6/x
    x=6/sina------(1)
    ---------------------
    <BAE=60-a
    sin(60-a)=9/x
    x=9/sin(60-a)--------(2)
    (1)=(2)
    6/sina=9/sin(60-a)
    sin(60-a)/sina=3/2
    sin60cota-cos60=3/2
    sin60cota=2
    tana=(sqrt3)/4
    sina=sqrt3/sqrt19
    sqrt3/sart19=6/x
    x=6sqrt19/sqrt3=2sqrt57

    ReplyDelete

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