【Haichuan Teahouse】Brain-teasing Series: Oddly-shaped final questions
Thread Content
In square ABCD, there is a semicircle and a quarter circle; the diameter of the semicircle is 7, and the radius of the quarter circle is 5. The common tangent to these two arcs intersects the square at points E and F. Find the length of the red segment EF.
The questions given by Yu are too difficult; I can’t solve them :(
The answer is 7/2+5=8.5; we will look at the details further tomorrow.
This post was last edited by chen/74 on 2021-7-28 at 20:20. Let the center of the semicircle be G, and let the side length of the square be L. Given that GD = a, draw line GC which intersects EF at the point of tangency H. Let EH = x and FH = y. By drawing line CE, we have 5/y = L/a and 8.5*8.5 = a*a + (a+3.5)(a+3.5); from these it follows that a = 4 and L = 7.5. Then (7.5 – x)*(7.5 – x) + 7.5*7.5 = x*x + 5*5, which gives CE*CE = x^2. Solving this yields x = 175/30 and y = 4/(7.5*5) = 40/15. Therefore, x + y = (175 + 80)/30 = 51/6 = 8.5. Additionally, by proving that the two perpendicular segments from the interior endpoints of the square to its sides are equal, it can be shown that 3.5 + 5 holds true. Using auxiliary lines that pass through the vertices of the square, it is possible to prove that the two triangles are congruent.
5/y = L/a, 8.5*8.5 = a*a + (a+3.5)(a+3.5); from these it can be deduced that a=4 and L=7.5. I don’t understand it – two equations with three unknowns, is it possible to arrive at a solution? If the values of a and L can be determined at this point, the value of y should also be available.
It is sufficient to prove that the two red triangles are congruent (the common tangent is the intersection of the line connecting the radii of the two circles after being rotated by 90 degrees with the two sides of the square)
Well, the approach is correct; I’ll give you 1 point, just in case you get proud. :Lol, the diagonals are equal, and the side lengths are equal as well.
It’s not difficult; just look at those who have already completed it upstairs. The graphical method can be used, and there are no complicated calculations involved – it’s based on simple principles
It’s a bit difficult to create a CAD diagram based on the information given in the problem; after much effort, the diagram was finally made and the solution could be found.