Prove that $2A+A_0=A_1+A_2+A_3$,where $A$ is the area of the triangle $ABC.$
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If $A_0$ denotes the area of the triangle formed by joining the points of contact of the inscribed circle of the triangle $ABC$ and the sides of the triangle;$A_1,A_2,A_3$ are the corresponding areas for the triangles thus formed with the escribed circles of the triangle $ABC.$Prove that $2A+A_0=A_1+A_2+A_3$,where $A$ is the area of the triangle $ABC.$
I found the area $A_0=A-frac12(s-a)^2-frac12(s-b)^2-frac12(s-c)^2$
Is my $A_0$ correct?
I am not able to find $A_1,A_2,A_3$.Please help.
geometry trigonometry
asked Apr 21 '16 at 8:27
Vinod Kumar Punia
2,634831
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If $A_0$ denotes the area of the triangle formed by joining the points of contact of the inscribed circle of the triangle $ABC$ and the sides of the triangle;$A_1,A_2,A_3$ are the corresponding areas for the triangles thus formed with the escribed circles of the triangle $ABC.$Prove that $2A+A_0=A_1+A_2+A_3$,where $A$ is the area of the triangle $ABC.$
I found the area $A_0=A-frac12(s-a)^2-frac12(s-b)^2-frac12(s-c)^2$
Is my $A_0$ correct?
I am not able to find $A_1,A_2,A_3$.Please help.
geometry trigonometry
asked Apr 21 '16 at 8:27
Vinod Kumar Punia
2,634831
can you be more explicit on how you define the "triangles thus formed with the escribed circles...". Is not clear...
– Chip
Apr 21 '16 at 8:30
The escribed circle opposite vertex $A$ touches $AB$ produced,$AC$ produced and $BC$,so $A_1$ is the area of triangle formed by points of contact of escribed circle with $BC,AB$ produced,$AC$ produced.Same for escribed circles opposite $B$ and $C.$
– Vinod Kumar Punia
Apr 21 '16 at 8:34
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
If $A_0$ denotes the area of the triangle formed by joining the points of contact of the inscribed circle of the triangle $ABC$ and the sides of the triangle;$A_1,A_2,A_3$ are the corresponding areas for the triangles thus formed with the escribed circles of the triangle $ABC.$Prove that $2A+A_0=A_1+A_2+A_3$,where $A$ is the area of the triangle $ABC.$
I found the area $A_0=A-frac12(s-a)^2-frac12(s-b)^2-frac12(s-c)^2$
Is my $A_0$ correct?
I am not able to find $A_1,A_2,A_3$.Please help.
geometry trigonometry
asked Apr 21 '16 at 8:27
Vinod Kumar Punia
2,634831
If $A_0$ denotes the area of the triangle formed by joining the points of contact of the inscribed circle of the triangle $ABC$ and the sides of the triangle;$A_1,A_2,A_3$ are the corresponding areas for the triangles thus formed with the escribed circles of the triangle $ABC.$Prove that $2A+A_0=A_1+A_2+A_3$,where $A$ is the area of the triangle $ABC.$
I found the area $A_0=A-frac12(s-a)^2-frac12(s-b)^2-frac12(s-c)^2$
Is my $A_0$ correct?
I am not able to find $A_1,A_2,A_3$.Please help.
geometry trigonometry
asked Apr 21 '16 at 8:27
Vinod Kumar Punia
2,634831
asked Apr 21 '16 at 8:27
Vinod Kumar Punia
2,634831
asked Apr 21 '16 at 8:27
Vinod Kumar Punia
2,634831
asked Apr 21 '16 at 8:27
Vinod Kumar Punia
2,634831
2,634831
can you be more explicit on how you define the "triangles thus formed with the escribed circles...". Is not clear...
– Chip
Apr 21 '16 at 8:30
The escribed circle opposite vertex $A$ touches $AB$ produced,$AC$ produced and $BC$,so $A_1$ is the area of triangle formed by points of contact of escribed circle with $BC,AB$ produced,$AC$ produced.Same for escribed circles opposite $B$ and $C.$
– Vinod Kumar Punia
Apr 21 '16 at 8:34
add a comment |Â
can you be more explicit on how you define the "triangles thus formed with the escribed circles...". Is not clear...
– Chip
Apr 21 '16 at 8:30
The escribed circle opposite vertex $A$ touches $AB$ produced,$AC$ produced and $BC$,so $A_1$ is the area of triangle formed by points of contact of escribed circle with $BC,AB$ produced,$AC$ produced.Same for escribed circles opposite $B$ and $C.$
– Vinod Kumar Punia
Apr 21 '16 at 8:34
can you be more explicit on how you define the "triangles thus formed with the escribed circles...". Is not clear...
– Chip
Apr 21 '16 at 8:30
can you be more explicit on how you define the "triangles thus formed with the escribed circles...". Is not clear...
– Chip
Apr 21 '16 at 8:30
The escribed circle opposite vertex $A$ touches $AB$ produced,$AC$ produced and $BC$,so $A_1$ is the area of triangle formed by points of contact of escribed circle with $BC,AB$ produced,$AC$ produced.Same for escribed circles opposite $B$ and $C.$
– Vinod Kumar Punia
Apr 21 '16 at 8:34
The escribed circle opposite vertex $A$ touches $AB$ produced,$AC$ produced and $BC$,so $A_1$ is the area of triangle formed by points of contact of escribed circle with $BC,AB$ produced,$AC$ produced.Same for escribed circles opposite $B$ and $C.$
– Vinod Kumar Punia
Apr 21 '16 at 8:34
add a comment |Â
2 Answers
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up vote
1
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Is my $A_0$ correct?
No, it is not correct. Let $D,E,F$ be the tangent point of the inscribed circle with the side $BC,CA,AB$ respectively. Then, noting that
$$AE=AF=s-a,quad BD=BF=s-b,quad CE=CD=s-c$$
where $s=(a+b+c)/2$, we have
$$beginalignA_0&=[triangleABC]-([triangleAEF]+[triangleBDF]+[triangleCDE])\&=A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin Cendaligntag1$$
By the way, let $K,L,M$ be the tangent point of the escribed circle in $angle A$ with the side $BC,CA,AB$ respectively. Then, noting that
$$CK=CL=s-b,quad BK=BM=s-c,$$
we have
$$beginalign[triangleKLM]&=[triangleAML]-([triangleABC]+[triangleBKM]+[triangleCKL])\&=frac 12s^2sin A-A-frac 12(s-c)^2sin(pi-B)-frac 12(s-b)^2sin(pi-C)\&=frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin Cendalign$$
Similarly, the areas of the other triangles formed with escribed circles are given by
$$frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A$$
$$frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A$$
Hence, from $(1)$, we have
$$2A+A_0-A_1-A_2-A_3$$$$=2A+A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin C-(frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin C)-(frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A)-(frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A)$$$$=6A+frac 12sin A(-(s-a)^2-s^2+(s-c)^2+(s-b)^2)$$$$+frac 12sin B(-(s-b)^2+(s-c)^2-s^2+(s-a)^2)$$$$+frac 12sin C(-(s-c)^2+(s-b)^2+(s-a)^2-s^2)$$$$=6A+frac 12sin A(-2bc)+frac 12sin B(-2ca)+frac 12sin C(-2ab)=6A-2A-2A-2A=0$$
answered Apr 21 '16 at 9:19
mathlove
87.1k877208
I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 10:52
1
@VinodKumarPunia: Since $a=BC=BK+KC=BM+CL$, we have $a+b+c=b+c+BM+CL=(b+CL)+(c+BM)=AL+AM$, and so we have $AL=AM=s$, from which $CK=CL=AL-AC=s-b$ and $BK=BM=AM-AB=s-c$ follow.
– mathlove
Apr 21 '16 at 11:11
How can we prove that $AE=AF=s-a,BD=BF=s-b,CD=CE=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 11:18
1
@VinodKumarPunia: Let $AE=AF=x,BD=BF=y,CD=CE=z$. Solving $z+y=a,x+z=b,x+y=c$ will give you the answer.
– mathlove
Apr 21 '16 at 11:24
I got it,thank you very much.@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 12:19
add a comment |Â
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0
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Hint:
$A_1=dfracA^22R(s-a)$ and $A_o=dfracA^22Rs$
Plug in the corresponding values in beginalign A_1+A_2+A_3-A_o&=dfracA^22Rcdotbigg[dfrac1s-a+dfrac1s-b+dfrac1s-c-dfrac1sbigg]\&=dfracA^22RcdotdfracabcA^2\&=dfrac2A^2cdot abc4Rcdot A^2\&=2A becauseA=fracabc4Rendalign
answered Aug 17 at 4:24
mnulb
1,325620
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Is my $A_0$ correct?
No, it is not correct. Let $D,E,F$ be the tangent point of the inscribed circle with the side $BC,CA,AB$ respectively. Then, noting that
$$AE=AF=s-a,quad BD=BF=s-b,quad CE=CD=s-c$$
where $s=(a+b+c)/2$, we have
$$beginalignA_0&=[triangleABC]-([triangleAEF]+[triangleBDF]+[triangleCDE])\&=A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin Cendaligntag1$$
By the way, let $K,L,M$ be the tangent point of the escribed circle in $angle A$ with the side $BC,CA,AB$ respectively. Then, noting that
$$CK=CL=s-b,quad BK=BM=s-c,$$
we have
$$beginalign[triangleKLM]&=[triangleAML]-([triangleABC]+[triangleBKM]+[triangleCKL])\&=frac 12s^2sin A-A-frac 12(s-c)^2sin(pi-B)-frac 12(s-b)^2sin(pi-C)\&=frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin Cendalign$$
Similarly, the areas of the other triangles formed with escribed circles are given by
$$frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A$$
$$frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A$$
Hence, from $(1)$, we have
$$2A+A_0-A_1-A_2-A_3$$$$=2A+A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin C-(frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin C)-(frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A)-(frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A)$$$$=6A+frac 12sin A(-(s-a)^2-s^2+(s-c)^2+(s-b)^2)$$$$+frac 12sin B(-(s-b)^2+(s-c)^2-s^2+(s-a)^2)$$$$+frac 12sin C(-(s-c)^2+(s-b)^2+(s-a)^2-s^2)$$$$=6A+frac 12sin A(-2bc)+frac 12sin B(-2ca)+frac 12sin C(-2ab)=6A-2A-2A-2A=0$$
answered Apr 21 '16 at 9:19
mathlove
87.1k877208
I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 10:52
1
@VinodKumarPunia: Since $a=BC=BK+KC=BM+CL$, we have $a+b+c=b+c+BM+CL=(b+CL)+(c+BM)=AL+AM$, and so we have $AL=AM=s$, from which $CK=CL=AL-AC=s-b$ and $BK=BM=AM-AB=s-c$ follow.
– mathlove
Apr 21 '16 at 11:11
How can we prove that $AE=AF=s-a,BD=BF=s-b,CD=CE=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 11:18
1
@VinodKumarPunia: Let $AE=AF=x,BD=BF=y,CD=CE=z$. Solving $z+y=a,x+z=b,x+y=c$ will give you the answer.
– mathlove
Apr 21 '16 at 11:24
I got it,thank you very much.@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 12:19
add a comment |Â
up vote
1
down vote
accepted
Is my $A_0$ correct?
No, it is not correct. Let $D,E,F$ be the tangent point of the inscribed circle with the side $BC,CA,AB$ respectively. Then, noting that
$$AE=AF=s-a,quad BD=BF=s-b,quad CE=CD=s-c$$
where $s=(a+b+c)/2$, we have
$$beginalignA_0&=[triangleABC]-([triangleAEF]+[triangleBDF]+[triangleCDE])\&=A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin Cendaligntag1$$
By the way, let $K,L,M$ be the tangent point of the escribed circle in $angle A$ with the side $BC,CA,AB$ respectively. Then, noting that
$$CK=CL=s-b,quad BK=BM=s-c,$$
we have
$$beginalign[triangleKLM]&=[triangleAML]-([triangleABC]+[triangleBKM]+[triangleCKL])\&=frac 12s^2sin A-A-frac 12(s-c)^2sin(pi-B)-frac 12(s-b)^2sin(pi-C)\&=frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin Cendalign$$
Similarly, the areas of the other triangles formed with escribed circles are given by
$$frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A$$
$$frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A$$
Hence, from $(1)$, we have
$$2A+A_0-A_1-A_2-A_3$$$$=2A+A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin C-(frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin C)-(frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A)-(frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A)$$$$=6A+frac 12sin A(-(s-a)^2-s^2+(s-c)^2+(s-b)^2)$$$$+frac 12sin B(-(s-b)^2+(s-c)^2-s^2+(s-a)^2)$$$$+frac 12sin C(-(s-c)^2+(s-b)^2+(s-a)^2-s^2)$$$$=6A+frac 12sin A(-2bc)+frac 12sin B(-2ca)+frac 12sin C(-2ab)=6A-2A-2A-2A=0$$
answered Apr 21 '16 at 9:19
mathlove
87.1k877208
I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 10:52
1
@VinodKumarPunia: Since $a=BC=BK+KC=BM+CL$, we have $a+b+c=b+c+BM+CL=(b+CL)+(c+BM)=AL+AM$, and so we have $AL=AM=s$, from which $CK=CL=AL-AC=s-b$ and $BK=BM=AM-AB=s-c$ follow.
– mathlove
Apr 21 '16 at 11:11
How can we prove that $AE=AF=s-a,BD=BF=s-b,CD=CE=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 11:18
1
@VinodKumarPunia: Let $AE=AF=x,BD=BF=y,CD=CE=z$. Solving $z+y=a,x+z=b,x+y=c$ will give you the answer.
– mathlove
Apr 21 '16 at 11:24
I got it,thank you very much.@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 12:19
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Is my $A_0$ correct?
No, it is not correct. Let $D,E,F$ be the tangent point of the inscribed circle with the side $BC,CA,AB$ respectively. Then, noting that
$$AE=AF=s-a,quad BD=BF=s-b,quad CE=CD=s-c$$
where $s=(a+b+c)/2$, we have
$$beginalignA_0&=[triangleABC]-([triangleAEF]+[triangleBDF]+[triangleCDE])\&=A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin Cendaligntag1$$
By the way, let $K,L,M$ be the tangent point of the escribed circle in $angle A$ with the side $BC,CA,AB$ respectively. Then, noting that
$$CK=CL=s-b,quad BK=BM=s-c,$$
we have
$$beginalign[triangleKLM]&=[triangleAML]-([triangleABC]+[triangleBKM]+[triangleCKL])\&=frac 12s^2sin A-A-frac 12(s-c)^2sin(pi-B)-frac 12(s-b)^2sin(pi-C)\&=frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin Cendalign$$
Similarly, the areas of the other triangles formed with escribed circles are given by
$$frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A$$
$$frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A$$
Hence, from $(1)$, we have
$$2A+A_0-A_1-A_2-A_3$$$$=2A+A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin C-(frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin C)-(frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A)-(frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A)$$$$=6A+frac 12sin A(-(s-a)^2-s^2+(s-c)^2+(s-b)^2)$$$$+frac 12sin B(-(s-b)^2+(s-c)^2-s^2+(s-a)^2)$$$$+frac 12sin C(-(s-c)^2+(s-b)^2+(s-a)^2-s^2)$$$$=6A+frac 12sin A(-2bc)+frac 12sin B(-2ca)+frac 12sin C(-2ab)=6A-2A-2A-2A=0$$
answered Apr 21 '16 at 9:19
mathlove
87.1k877208
Is my $A_0$ correct?
No, it is not correct. Let $D,E,F$ be the tangent point of the inscribed circle with the side $BC,CA,AB$ respectively. Then, noting that
$$AE=AF=s-a,quad BD=BF=s-b,quad CE=CD=s-c$$
where $s=(a+b+c)/2$, we have
$$beginalignA_0&=[triangleABC]-([triangleAEF]+[triangleBDF]+[triangleCDE])\&=A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin Cendaligntag1$$
By the way, let $K,L,M$ be the tangent point of the escribed circle in $angle A$ with the side $BC,CA,AB$ respectively. Then, noting that
$$CK=CL=s-b,quad BK=BM=s-c,$$
we have
$$beginalign[triangleKLM]&=[triangleAML]-([triangleABC]+[triangleBKM]+[triangleCKL])\&=frac 12s^2sin A-A-frac 12(s-c)^2sin(pi-B)-frac 12(s-b)^2sin(pi-C)\&=frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin Cendalign$$
Similarly, the areas of the other triangles formed with escribed circles are given by
$$frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A$$
$$frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A$$
Hence, from $(1)$, we have
$$2A+A_0-A_1-A_2-A_3$$$$=2A+A-frac 12(s-a)^2sin A-frac 12(s-b)^2sin B-frac 12(s-c)^2sin C-(frac 12s^2sin A-A-frac 12(s-c)^2sin B-frac 12(s-b)^2sin C)-(frac 12s^2sin B-A-frac 12(s-a)^2sin C-frac 12(s-c)^2sin A)-(frac 12s^2sin C-A-frac 12(s-a)^2sin B-frac 12(s-b)^2sin A)$$$$=6A+frac 12sin A(-(s-a)^2-s^2+(s-c)^2+(s-b)^2)$$$$+frac 12sin B(-(s-b)^2+(s-c)^2-s^2+(s-a)^2)$$$$+frac 12sin C(-(s-c)^2+(s-b)^2+(s-a)^2-s^2)$$$$=6A+frac 12sin A(-2bc)+frac 12sin B(-2ca)+frac 12sin C(-2ab)=6A-2A-2A-2A=0$$
answered Apr 21 '16 at 9:19
mathlove
87.1k877208
edited Apr 21 '16 at 10:45
answered Apr 21 '16 at 9:19
mathlove
87.1k877208
answered Apr 21 '16 at 9:19
mathlove
87.1k877208
answered Apr 21 '16 at 9:19
mathlove
87.1k877208
87.1k877208
I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 10:52
1
@VinodKumarPunia: Since $a=BC=BK+KC=BM+CL$, we have $a+b+c=b+c+BM+CL=(b+CL)+(c+BM)=AL+AM$, and so we have $AL=AM=s$, from which $CK=CL=AL-AC=s-b$ and $BK=BM=AM-AB=s-c$ follow.
– mathlove
Apr 21 '16 at 11:11
How can we prove that $AE=AF=s-a,BD=BF=s-b,CD=CE=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 11:18
1
@VinodKumarPunia: Let $AE=AF=x,BD=BF=y,CD=CE=z$. Solving $z+y=a,x+z=b,x+y=c$ will give you the answer.
– mathlove
Apr 21 '16 at 11:24
I got it,thank you very much.@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 12:19
add a comment |Â
I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 10:52
1
@VinodKumarPunia: Since $a=BC=BK+KC=BM+CL$, we have $a+b+c=b+c+BM+CL=(b+CL)+(c+BM)=AL+AM$, and so we have $AL=AM=s$, from which $CK=CL=AL-AC=s-b$ and $BK=BM=AM-AB=s-c$ follow.
– mathlove
Apr 21 '16 at 11:11
How can we prove that $AE=AF=s-a,BD=BF=s-b,CD=CE=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 11:18
1
@VinodKumarPunia: Let $AE=AF=x,BD=BF=y,CD=CE=z$. Solving $z+y=a,x+z=b,x+y=c$ will give you the answer.
– mathlove
Apr 21 '16 at 11:24
I got it,thank you very much.@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 12:19
I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 10:52
I do not understand how $CK=CL=s-b$ and $BK=BM=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 10:52
1
1
@VinodKumarPunia: Since $a=BC=BK+KC=BM+CL$, we have $a+b+c=b+c+BM+CL=(b+CL)+(c+BM)=AL+AM$, and so we have $AL=AM=s$, from which $CK=CL=AL-AC=s-b$ and $BK=BM=AM-AB=s-c$ follow.
– mathlove
Apr 21 '16 at 11:11
@VinodKumarPunia: Since $a=BC=BK+KC=BM+CL$, we have $a+b+c=b+c+BM+CL=(b+CL)+(c+BM)=AL+AM$, and so we have $AL=AM=s$, from which $CK=CL=AL-AC=s-b$ and $BK=BM=AM-AB=s-c$ follow.
– mathlove
Apr 21 '16 at 11:11
How can we prove that $AE=AF=s-a,BD=BF=s-b,CD=CE=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 11:18
How can we prove that $AE=AF=s-a,BD=BF=s-b,CD=CE=s-c$?@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 11:18
1
1
@VinodKumarPunia: Let $AE=AF=x,BD=BF=y,CD=CE=z$. Solving $z+y=a,x+z=b,x+y=c$ will give you the answer.
– mathlove
Apr 21 '16 at 11:24
@VinodKumarPunia: Let $AE=AF=x,BD=BF=y,CD=CE=z$. Solving $z+y=a,x+z=b,x+y=c$ will give you the answer.
– mathlove
Apr 21 '16 at 11:24
I got it,thank you very much.@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 12:19
I got it,thank you very much.@mathlove
– Vinod Kumar Punia
Apr 21 '16 at 12:19
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Hint:
$A_1=dfracA^22R(s-a)$ and $A_o=dfracA^22Rs$
Plug in the corresponding values in beginalign A_1+A_2+A_3-A_o&=dfracA^22Rcdotbigg[dfrac1s-a+dfrac1s-b+dfrac1s-c-dfrac1sbigg]\&=dfracA^22RcdotdfracabcA^2\&=dfrac2A^2cdot abc4Rcdot A^2\&=2A becauseA=fracabc4Rendalign
answered Aug 17 at 4:24
mnulb
1,325620
add a comment |Â
up vote
0
down vote
Hint:
$A_1=dfracA^22R(s-a)$ and $A_o=dfracA^22Rs$
Plug in the corresponding values in beginalign A_1+A_2+A_3-A_o&=dfracA^22Rcdotbigg[dfrac1s-a+dfrac1s-b+dfrac1s-c-dfrac1sbigg]\&=dfracA^22RcdotdfracabcA^2\&=dfrac2A^2cdot abc4Rcdot A^2\&=2A becauseA=fracabc4Rendalign
answered Aug 17 at 4:24
mnulb
1,325620
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
$A_1=dfracA^22R(s-a)$ and $A_o=dfracA^22Rs$
Plug in the corresponding values in beginalign A_1+A_2+A_3-A_o&=dfracA^22Rcdotbigg[dfrac1s-a+dfrac1s-b+dfrac1s-c-dfrac1sbigg]\&=dfracA^22RcdotdfracabcA^2\&=dfrac2A^2cdot abc4Rcdot A^2\&=2A becauseA=fracabc4Rendalign
answered Aug 17 at 4:24
mnulb
1,325620
Hint:
$A_1=dfracA^22R(s-a)$ and $A_o=dfracA^22Rs$
Plug in the corresponding values in beginalign A_1+A_2+A_3-A_o&=dfracA^22Rcdotbigg[dfrac1s-a+dfrac1s-b+dfrac1s-c-dfrac1sbigg]\&=dfracA^22RcdotdfracabcA^2\&=dfrac2A^2cdot abc4Rcdot A^2\&=2A becauseA=fracabc4Rendalign
answered Aug 17 at 4:24
mnulb
1,325620
edited Aug 17 at 4:39
answered Aug 17 at 4:24
mnulb
1,325620
answered Aug 17 at 4:24
mnulb
1,325620
answered Aug 17 at 4:24
mnulb
1,325620
1,325620
add a comment |Â
add a comment |Â
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can you be more explicit on how you define the "triangles thus formed with the escribed circles...". Is not clear...
– Chip
Apr 21 '16 at 8:30
The escribed circle opposite vertex $A$ touches $AB$ produced,$AC$ produced and $BC$,so $A_1$ is the area of triangle formed by points of contact of escribed circle with $BC,AB$ produced,$AC$ produced.Same for escribed circles opposite $B$ and $C.$
– Vinod Kumar Punia
Apr 21 '16 at 8:34