【Haichuan Teahouse】Brain-teaser Series: How Can Four Triangles Be Equivalent?
Thread Content
As shown in the figure, the side length of square ABCD is 8, with AE = CF = 2. Connect EF and fold quadrilateral ABFE along EF; what is the area of the shaded region? The key to this problem is to prove that the four small triangles are congruent, but after thinking about it for a day, I was only able to show that triangles 1 and 4, as well as triangles 2 and 3, are congruent respectively.
The 【Haichuan Equipment Guessing】 series is being updated continuously in 2021; all Haiyu friends are welcome to participate actively! https://bbs.hcbbs.com/thread-2079686-1-1.html (Source: Haichuan Chemical Industry Forum – Huahai Chuanliu) ------------------------------------------------------------------------------- A compilation post in the 【Haichuan Safety Discussion】 series, which is being updated continuously; everyone is welcome to participate in the discussions! https://bbs.hcbbs.com/thread-3235104-1-1.html (Source: Haichuan Chemical Industry Forum – Huahai Chuanliu)
This post was last edited by Thank goodness you're here on 2021-7-12 09:28 with WTF. I started two posts at the same time and ended up replying to the wrong one. Does the advisory team no longer have the authority to delete posts?
Let the area of the shadow be S. Then, 2S + 2*4 = 64, so S = 28
The answer should be 26.6666……6. So, your reply of 666 is quite meaningful; P
Calculated using a crude method: 1. It’s certain that S1 = S4; in a right triangle, the area is always the same as the product of the lengths of its legs； 2. By folding the square back along EF, it can be deduced that b + ax = 6; x is a multiple of the similarity ratio, and thus S2 = S3 ； 3. Next, to prove that S1 = S2, I calculated the length of the hypotenuse of the triangle and then compared them; it’s a clumsy method: I wrote down 3 equations: a^2 + 4 = b^2 ； b+ax = 6 ； a+bx+2x = 8. The calculations yield b=10/3 and a=8/3; ax=6-b=6-10/3=8/3=a. The multiplier x is 1, so the areas are equal. The calculation steps are rather complicated: (1) From b+ax=6, we get a=(6-b)/x ； (2) Substituting into the formula a^2+4-b^2=(6-b)^2/x^2+4-b^2=0, we get x^2=(6-b)^2/(b^2-4) ； (3) By substituting (1) into the equation a+bx+2x=(6-b)/x+(b+2)x=8, we get (6-b)+(b+2)x^2-8x=0. This can be rewritten as (6-b)+(b+2)*(6-b)^2/(b^2-4)-8x=0. Solving for x gives x=(6-b)/2(b-2). Squaring this value yields the same result as in (2), namely x^2=(6-b)^2/4(b-2)^2=(6-b)^2/(b^2-4). From this, we have 4(b-2)^2=b^2-4; solving this equation gives 4(b-2)=b+2, so 3b=10 and therefore b=10/3 ； (4) By substituting b=10/3 into the equation above, we obtain a=8/3. Then ax=6-b=6-10/3=8/3, which is still equal to a; therefore, the areas are all the same.
Triangle, with sides of 2=6/3, 8/3, and 10/3; area = 2*8/3 /2 = 8/3.
This post was last edited by Yang Qing on 2021-7-12 at 11:37. Thus, the area of the shadow = (area of the trapezoid – 2*8/3) = 26.6666666666……6; P is the correct answer. It seems a bit complicated, but there doesn’t seem to be any better solution. Note: The so-called trapezoid is shown in the diagram on floor 6.