Finding area of successive triangles
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Let the lines $L_1equivsqrt{3} x-y +(2-sqrt{3})=0$ and $L_2equivsqrt{3} x+y -(2+sqrt{3})=0$ intersect at $A$. Let $B_1$ be a point on $L_1$. From $B_1$, draw a line perpendicular to $L_1$ meeting the line $L_2$ at $C_1$. From $C_1$, draw a line perpendicular to $L_2$ meeting the line $L_1$ at $B_2$. Continue in this way obtaining points $C_2$, $B_3$, $C_3$ and so on. If area of triangle $AB_1C_1=1$ and $text{area}(Delta AB_2C_2) +text{area}(Delta AB_3C_3) +text{area}(Delta AB_4C_4) =T$, then $|T-4360|={}$_________________.
In this question I tried using co-ordinate plane.
For the first triangle I could get $B_1$ but the coordinates are
very clumsy.
Is there some geometrical method?
geometry
edited Nov 18 at 5:45
Arthur
108k7103186
asked Nov 18 at 5:29
maveric
61611
|
show 3 more comments
up vote
0
down vote
favorite
Let the lines $L_1equivsqrt{3} x-y +(2-sqrt{3})=0$ and $L_2equivsqrt{3} x+y -(2+sqrt{3})=0$ intersect at $A$. Let $B_1$ be a point on $L_1$. From $B_1$, draw a line perpendicular to $L_1$ meeting the line $L_2$ at $C_1$. From $C_1$, draw a line perpendicular to $L_2$ meeting the line $L_1$ at $B_2$. Continue in this way obtaining points $C_2$, $B_3$, $C_3$ and so on. If area of triangle $AB_1C_1=1$ and $text{area}(Delta AB_2C_2) +text{area}(Delta AB_3C_3) +text{area}(Delta AB_4C_4) =T$, then $|T-4360|={}$_________________.
In this question I tried using co-ordinate plane.
For the first triangle I could get $B_1$ but the coordinates are
very clumsy.
Is there some geometrical method?
geometry
edited Nov 18 at 5:45
Arthur
108k7103186
asked Nov 18 at 5:29
maveric
61611
Hint: all the triangles $AB_iC_i$ are similar, and their areas follow a geometric progression. I think working with angles and lengths is easier than coordinates, but I haven't actually tried.
– Arthur
Nov 18 at 5:37
But for finding the ratio wont we require use of coordinate geometry.
– maveric
Nov 18 at 5:44
Maybe. As I said, I haven't tried.
– Arthur
Nov 18 at 5:45
how do they follow gp?
– maveric
Nov 18 at 5:46
1
Because the figures $AB_iC_iB_{i+1}C_{i+1}$ are all similar. So the ratio $frac{|AB_{i+1}C_{i+1}|}{|AB_iC_i|}$ of areas must be constant.
– Arthur
Nov 18 at 5:54
|
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let the lines $L_1equivsqrt{3} x-y +(2-sqrt{3})=0$ and $L_2equivsqrt{3} x+y -(2+sqrt{3})=0$ intersect at $A$. Let $B_1$ be a point on $L_1$. From $B_1$, draw a line perpendicular to $L_1$ meeting the line $L_2$ at $C_1$. From $C_1$, draw a line perpendicular to $L_2$ meeting the line $L_1$ at $B_2$. Continue in this way obtaining points $C_2$, $B_3$, $C_3$ and so on. If area of triangle $AB_1C_1=1$ and $text{area}(Delta AB_2C_2) +text{area}(Delta AB_3C_3) +text{area}(Delta AB_4C_4) =T$, then $|T-4360|={}$_________________.
In this question I tried using co-ordinate plane.
For the first triangle I could get $B_1$ but the coordinates are
very clumsy.
Is there some geometrical method?
geometry
edited Nov 18 at 5:45
Arthur
108k7103186
asked Nov 18 at 5:29
maveric
61611
Let the lines $L_1equivsqrt{3} x-y +(2-sqrt{3})=0$ and $L_2equivsqrt{3} x+y -(2+sqrt{3})=0$ intersect at $A$. Let $B_1$ be a point on $L_1$. From $B_1$, draw a line perpendicular to $L_1$ meeting the line $L_2$ at $C_1$. From $C_1$, draw a line perpendicular to $L_2$ meeting the line $L_1$ at $B_2$. Continue in this way obtaining points $C_2$, $B_3$, $C_3$ and so on. If area of triangle $AB_1C_1=1$ and $text{area}(Delta AB_2C_2) +text{area}(Delta AB_3C_3) +text{area}(Delta AB_4C_4) =T$, then $|T-4360|={}$_________________.
In this question I tried using co-ordinate plane.
For the first triangle I could get $B_1$ but the coordinates are
very clumsy.
Is there some geometrical method?
geometry
geometry
edited Nov 18 at 5:45
Arthur
108k7103186
asked Nov 18 at 5:29
maveric
61611
edited Nov 18 at 5:45
Arthur
108k7103186
asked Nov 18 at 5:29
maveric
61611
edited Nov 18 at 5:45
Arthur
108k7103186
edited Nov 18 at 5:45
Arthur
108k7103186
edited Nov 18 at 5:45
Arthur
108k7103186
108k7103186
asked Nov 18 at 5:29
maveric
61611
asked Nov 18 at 5:29
maveric
61611
asked Nov 18 at 5:29
maveric
61611
61611
Hint: all the triangles $AB_iC_i$ are similar, and their areas follow a geometric progression. I think working with angles and lengths is easier than coordinates, but I haven't actually tried.
– Arthur
Nov 18 at 5:37
But for finding the ratio wont we require use of coordinate geometry.
– maveric
Nov 18 at 5:44
Maybe. As I said, I haven't tried.
– Arthur
Nov 18 at 5:45
how do they follow gp?
– maveric
Nov 18 at 5:46
1
Because the figures $AB_iC_iB_{i+1}C_{i+1}$ are all similar. So the ratio $frac{|AB_{i+1}C_{i+1}|}{|AB_iC_i|}$ of areas must be constant.
– Arthur
Nov 18 at 5:54
|
show 3 more comments
Hint: all the triangles $AB_iC_i$ are similar, and their areas follow a geometric progression. I think working with angles and lengths is easier than coordinates, but I haven't actually tried.
– Arthur
Nov 18 at 5:37
But for finding the ratio wont we require use of coordinate geometry.
– maveric
Nov 18 at 5:44
Maybe. As I said, I haven't tried.
– Arthur
Nov 18 at 5:45
how do they follow gp?
– maveric
Nov 18 at 5:46
1
Because the figures $AB_iC_iB_{i+1}C_{i+1}$ are all similar. So the ratio $frac{|AB_{i+1}C_{i+1}|}{|AB_iC_i|}$ of areas must be constant.
– Arthur
Nov 18 at 5:54
Hint: all the triangles $AB_iC_i$ are similar, and their areas follow a geometric progression. I think working with angles and lengths is easier than coordinates, but I haven't actually tried.
– Arthur
Nov 18 at 5:37
Hint: all the triangles $AB_iC_i$ are similar, and their areas follow a geometric progression. I think working with angles and lengths is easier than coordinates, but I haven't actually tried.
– Arthur
Nov 18 at 5:37
But for finding the ratio wont we require use of coordinate geometry.
– maveric
Nov 18 at 5:44
But for finding the ratio wont we require use of coordinate geometry.
– maveric
Nov 18 at 5:44
Maybe. As I said, I haven't tried.
– Arthur
Nov 18 at 5:45
Maybe. As I said, I haven't tried.
– Arthur
Nov 18 at 5:45
how do they follow gp?
– maveric
Nov 18 at 5:46
how do they follow gp?
– maveric
Nov 18 at 5:46
1
1
Because the figures $AB_iC_iB_{i+1}C_{i+1}$ are all similar. So the ratio $frac{|AB_{i+1}C_{i+1}|}{|AB_iC_i|}$ of areas must be constant.
– Arthur
Nov 18 at 5:54
Because the figures $AB_iC_iB_{i+1}C_{i+1}$ are all similar. So the ratio $frac{|AB_{i+1}C_{i+1}|}{|AB_iC_i|}$ of areas must be constant.
– Arthur
Nov 18 at 5:54
|
show 3 more comments
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Hint: all the triangles $AB_iC_i$ are similar, and their areas follow a geometric progression. I think working with angles and lengths is easier than coordinates, but I haven't actually tried.
– Arthur
Nov 18 at 5:37
But for finding the ratio wont we require use of coordinate geometry.
– maveric
Nov 18 at 5:44
Maybe. As I said, I haven't tried.
– Arthur
Nov 18 at 5:45
how do they follow gp?
– maveric
Nov 18 at 5:46
1
Because the figures $AB_iC_iB_{i+1}C_{i+1}$ are all similar. So the ratio $frac{|AB_{i+1}C_{i+1}|}{|AB_iC_i|}$ of areas must be constant.
– Arthur
Nov 18 at 5:54