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Math Help - Squares in triangles ?

  1. #1
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    Question Squares in triangles ?

    Hi everyone !

    I've got some trouble...

    IJKL is a square and [IJ] is confunded with [BC]
    h is the height from A.
    C1 is the red square
    C2 is the green square
    C3 is the blue square
    a= [BC]
    b= [CA]
    c= [AB]

    1. How to find IJKL using a and h
    2. a< b< c. Classify C1, C2, C3 from the largest.
    3. Find ABC triangles (their surface, S) when the square IJKL has the largest surface.

    Thank you for your help.
    Attached Thumbnails Attached Thumbnails Squares in triangles ?-square-triangle.jpg  
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  2. #2
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    Re: Squares in triangles ?

    Working on parts 2 & 3, but I figured I would post part 1. In your drawing, LK and BC are parallel, making \angleALK = \angleABC and \angleAKL = \angleACB. Therefore, \triangleALK is similar to \triangleABC. Since these triangles are similar:

    (1) \frac{LK}{BC}\ =\ \frac{AK}{AC}

    If the side length of the inscribed square above side a is s, then the equation above says:

    (2) \frac{s}{a}\ =\ \frac{AK}{AC}

    Now we draw the height of the triangle down from vertex A to side a, and call it h, meeting at the point H on a. Let the intersection of AH and LK be D. Line LK splits AH into two segments, the upper (AD) which has length h-s and the lower (DH) which has length s. Then, we now know that \triangleADK is similar to \triangleAHC. Therefore,

    (3) \frac{AK}{AC}\ =\ \frac{AD}{AH}

    And,

    (4) \frac{AD}{AH}\ =\ \frac{h-s}{h}

    Thus, by equations (2), (3), and (4),

    (5) \frac{s}{a}\ =\ \frac{h-s}{h}

    And then, solving for s, gives:

    (6) s\ =\ \frac{ah}{a+h}

    Could you explain exactly what you mean in part 3. of your question a little more? Thanks.
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  3. #3
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    Re: Squares in triangles ?

    Alright, so I think I've figured out part two. To start, draw the altitudes to each of the sides in the acute \triangleABC. Let the the altitude to side a be ha, the altitude to side b be hb, and the altitude to side c be hc. Then after a few seconds looking at the many right triangles created, I think you will agree that:

    (1) \sin(A)\ =\ \frac{h_{b}}{c}\ =\ \frac{h_{c}}{b}

    (2) \sin(B)\ =\ \frac{h_{c}}{a}\ =\ \frac{h_{a}}{c}

    (3) \sin(C)\ =\ \frac{h_{a}}{b}\ =\ \frac{h_{b}}{a}

    From these equations, it follows that:

    (4) ah_{a}\ =\ bh_{b}\ =\ ch_{c}

    And we know that, from my previous post:

    C_{1}\ =\ \frac{ah_{a}}{a+h_{a}}

    C_{2}\ =\ \frac{bh_{b}}{b+h_{b}}

    C_{3}\ =\ \frac{ch_{c}}{c+h_{c}}

    We will show that C_{3}\ <\ C_{2}\ <\ C_{1}.

    First, we know that \sin(C)\ <\ 1 because C is a positive angle in an acute triangle. Then we multiply both sides of this inequality by the positive quantity (b-a) to get:

    (b-a)\sin(C)\ <\ (b-a)

    b\sin(C)-a\sin(C)\ <\ b-a

    By (3), b\sin(C) = h_{a} and a\sin(C) = h_{b}. So:

    h_{a}-h_{b}\ <\ b-a

    a+h_{a}\ <\ b+h_{b}

    \frac{1}{a+h_{a}}\ >\ \frac{1}{b+h_{b}}

    And since, by (4), ah_{a}\ =\ bh_{b},

    \frac{ah_{a}}{a+h_{a}}\ >\ \frac{bh_{b}}{b+h_{b}}

    i.e.

    C1 > C2. A similar procedure shows that C2 > C3. Thus, we have shown that:

    C_{3}\ <\ C_{2}\ <\ C_{1}
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  4. #4
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    Re: Squares in triangles ?

    For part three, this problem is similar to minimizing the perimeter of a square while keeping the area constant. We know that ah_{a} is constant because it is twice the area of the triangle (i.e. 2 times a constant A). Thus, we just have to minimize the quantity a+h_{a} to maximize C1. Since the area is constant, we have that:

    \frac{ah_{a}}{2}\ =\ A

    h_{a}\ =\ \frac{2A}{a}

    Thus, our goal is to minimize:

    f(a)\ =\ a+\frac{2A}{a}

    One way to do this is to treat it as a function of a on the domain [0, \infty). This is a differentiable function on (0, \infty), so if it has a minimum it will be at a place where the derivative is zero. i.e.

    f'(a)\ =\ 1-\frac{2A}{a^2}\ =\ 0

    Solving for a gives,

    a\ =\ \sqrt{2A}

    This is a global min for f on [0, \infty). And so:

    h_{a}\ =\ \frac{2A}{a}\ =\ \frac{2A}{\sqrt{2A}}\ =\ \sqrt{2A}

    Thus,

    C_{1}\ =\ \frac{\sqrt{2A}^2}{\sqrt{2A}+\sqrt{2A}}\ =\ \frac{\sqrt{2A}}{2}

    And so, in conclusion, the side length of the square is maximized when a\ =\ h_{a}\ =\ \sqrt{2A}. When this happens, the area of the inscribed square is half the area of the triangle. As far as the type of triangle in which this can happen, there is a little choice. It can be an isosceles right, equilateral, or scalene triangle. To see this, draw a square of length a. The base of the square is the side BC and A can be any point on the top line of the square. Drawing the triangle between these three points gives a triangle with a\ =\ h_{a}\ =\ \sqrt{2A}, and so the C1 square inside any one of these triangles is the same size.
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