# Squares in a triangle?

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.

-
What is this??. – DonAntonio Nov 30 '12 at 20:23
You may want to upload the picture somewhere else, because I have to register to Math Help Forum to be able to see the picture to which you are linking. – anonymous Nov 30 '12 at 20:23
Can you see the picture ? – Dan Nov 30 '12 at 20:38
@Dan, I separated the statements as you did in your post (use two "enters" to separate lines, or when you want to begin another line, use <br>. I assumed that, e.g., [BC] means the length |BC|? I'm not clear what you mean by [IJ] is confunded with [BC]. Do you mean confounded? Or that |IJ| is proportional to |BC|, and varies as |BC| varies? – amWhy Nov 30 '12 at 20:49
In fact, I want to express the lenght of the sides of IJKL using a and h. Did you understand ? – Dan Nov 30 '12 at 21:06

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 $C_3$.
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.