Browsed on the Internet to a "super puzzle",
Figure 1. Original question:As shown in Figure 2,
Figure II. The square is divided into 5 regions by AE and CD, ab=4, bc=6, ab perpendicular bc, and the sum of the shaded part area is found.
Unfortunately, this is a mistake!
1. The "mistakes" and analysis of the original question
Rotate AFE 90° clockwise around point E to coincide with F and denote it as A'CE, point A'On the extension line on the left side of the square, as shown below.
Figure III. Observe.
BCE+ BEC=90°= AEF+ BEC, so AEF= BCE. Again'EC is the result of AEF rotation, hence A'ec = aef = bce, thus dc a'e。Also known de a'c, hence the quadrilateral a'CDE is a parallelogram.
From the parallelogram to the equal sides, ae=a'e=cd。
Remembering bd=x, then from ae=bd, ab=4, and bc=6, we can know be=x+2, as shown in the figure.
Figure IV. From the projective theorem, we know that be = bd bc, that is, (x+2) = 6x. Simplification yields x -2x + 4 = 0, i.e. (x-1) + 3 = 0, contradiction!So BD has no solution, iePoint b is not on the CD
Therefore point b satisfying Figure 1 does not existThis is where the mistake of the original question lies
2. The original question is wrong, but it does not affect the answer!Although the answer is available, it is only a "castle in the sky"!
by ".Error analysis"It can be seen that s aef=s cde.
From "the area of the equal-bottom contour triangle is equal", it can be seen that S ace=S AEF+S ACG, as shown in Figure 5.
Figure V. It is obtained from and .
s ace = s aef + s acg = s cde + s acg, i.e. .
s abc+s bce=s bce+s bde+s acg, thus.
s abc = s bde + s acg = s shade.
s shade = s abc = 4 6 2 = 12.
3. Correction of the original question
Place the original question condition".ab=4，bc=6Modified toab=3，bc=4", as shown in Fig.
Figure VI. Same as ".Analysis of errors in the original question", bd is x, then be=x+1, from the projective theorem (x+1) =4x, to find the unique solution x=1,That is, point b satisfies the conditions of Figure 6 and is the only one!Therefore, the revised question is correct!
Same as ".II. II. II, we can find s shadow = s abc = 6.
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