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Math problems in the second day of junior high school (math report)
p= 1/2[a+b+c]

a+b+c=2p

1/4[a^2b^2-( 1/2(a^2+b^2-c^2))^2]

= 1/4[ab+ 1/2(a^2+b^2-c^2)][ab- 1/2(a^2+b^2-c^2]

= 1/ 16[2ab+a^2+b^2-c^2][2ab- a^2-b^2+c^2]

= 1/ 16[(a+b)^2-c^2][c^2-(a-b)^2]

= 1/ 16[a+b-c][a+b+c][c-a+b][c+a-b]

= 1/ 16[a+b+ c-2c][a+b+ c][c-2a+b+ a][c+a+B- 2b]

= 1/ 16[2p-2c]2p[2p-2a][2p-2b]

=[p-c]p[p-a][p-b]

=p[p-a][p-b] [p-c]

Namely:

1/4[a^2b^2-( 1/2(a^2+b^2-c^2))^2]=p[p-a][p-b]