InterviewSolution
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InterviewSolution
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Let the incircle of a Delta ABC touches sides BC, CA and AB at D,E and F, respectively. Let area of Delta ABC be Delta and thatof DEF be Delta'. If a, b and c are side of Dela ABC, then the value of abc(a+b+c)(Delta')/(Delta^(3)) is |
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Answer» 1  Clearly points C, D, I and E are concyclic `therefore angle EID = pi - C` `therefore` In `Delta DEF, angle DFE =(pi-C)/(2)` Similarly `angle EDF =(pi-A)/(2)` and `angle FED = (pi-B)/(2)` `Delta' =2R^(2)SIN((pi-A)/(2))sin((pi-B)/(2))sin((pi-C)/(2))` `=2r^(2)cos((A)/(2))cos((B)/(2))cos((C )/(2))` `therefore abc (a+b+c)(Delta')/(Delta^(3))` `=(abc(a+b+c))/(Delta^(3))2r^(2)cos.(A)/(2)cos.(B)/(2)cos.(C )/(2)` `=(abc(2s))/(Delta^(3))2r^(2)cos.(A)/(2)cos.(B)/(2)cos.(C )/(2)` `=(abc(2s))/(Delta^(3))2r^(2)sqrt((s(s-a))/(bc))sqrt((s(s-b))/(ab))sqrt((s(s-c))/(ac))` `=(4abcs)/(Delta^(3))r^(2)(s)/(abc)sqrt(s(s-a)(s-b)(s-c))` `=(4S^(2))/(Delta^(2))r^(2)` = 4. |
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