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

4 comments:

  1. 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.

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  2. 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

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  3. 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

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  4. 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