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这道题蛮有趣的,试解一下,期待高手给出更简洁的解法

monseigneur 2024-03-03 19:20:07 ( reads)

Let bottom of triangle be b, height be h. Let the whole triangle area be S = bh/2. Assume S=1 for now. Mark the area of the small triangle on each corner as x1, x2, x3. Mark the area in question as x. Obviously, x = x1+x2+x3.

Draw BC in parallel with DE. ABC area = (7/9)b*(2/3)h/2=14/27.  ADE area = ABC*((6/7)^2)=8/21.

Similarly, area of  BFG, HIJ are also the same value. These three triangles piece together to form the large triangle, except that some areas are uncovered, some are overlapping. By making necessary adjustment, we have:

3*ADE=1+2x-(x1+x2+x3)=1+2x-x=1+x

x=3*ADE-1=3*8/21-1=1/7

If S=35, x=S*(1/7)=5
 

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