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I've got some trouble...

IJKL is a square and B, I, J, C are aligned (alternatively, |IJ| is confounded with |BC|.

h is the height of acute $\triangle$ ABC from A to side BC.

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 area, S) when the square IJKL has the largest area.

squares in triangle

Thank you in advance.

  • 0
    I updated my answer with a trigonometric proof of 2., though I would still like a geometric one. The triangle being acute isn't as crucial as I thought--it just can't be a right triangle.2012-11-30

1 Answers 1

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  1. Let $s = |IJ|$ be the side length of the square. Consider the similar triangles $ABC$ and $ALK$. The ratio of the height of $ALK$ to the height of $ABC$ is the same as the ratio of the base of $ALK$ to the base of $ABC$. That is, $\frac{h-s}{h} = \frac{s}{a}.$ Cross multiply to obtain $ah - as = sh$, and solve for $s$ to obtain $s = \frac{ah}{a+h}.$

  2. Let $h_a = h$ be the height when we consider $a$ to be the base, and define $h_b$ and $h_c$ correspondingly. Then from the first part, the side length of $C_1$ is $\frac{ah_a}{a+h_a}$, for $C_2$ it is $\frac{bh_b}{b+h_b}$, and for $C_3$ it is $\frac{ch_c}{c+h_c}$. The numerator is constant in each case (because it is proportional to the height of the whole triangle), so we need only consider the denominator. Note that we can write $h_a = b\sin C = c \sin B$, and the other two heights analogously. Then $(a + h_a) - (b + h_b) = a + b\sin C - b - a\sin C = (a-b)(1 - \sin C) < 0,$ since $a < b$ and $\sin C < 1$. Thus $a+h_a < b+h_b$, and so the side length of $C_1$ is larger than the side length of $C_2$. By exactly the same method we can prove that $C_2$ is larger than $C_3$, so from largest to smallest we have $C_1$, $C_2$, and then $C_3$.
    For this to hold we actually don't need the triangle to be acute, we just need it not to be a right triangle.

  3. The area of $IJKL$ will be maximized relative to the total area $S$ of the triangle when $a = h$, since this minimizes $a+h$ for fixed $ah$. In this case the area of the square is exactly one-half the area of the triangle.

  • 2
    Similar triangles are one of the most basic and important ideas in plane geometry. There's probably another way to prove it without similar triangles, but almost certainly not as elegant. I suggest you learn what similar triangles are so that you can use them.2012-11-30