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The area of a triangle determined by the bisectors.
$begingroup$
How can I calculate the area of a triangle determined by the interior bisectors? What I want to say it is represented in the following picture:
$AQ$ is the bisector of the angle $angle BAC$, $BR$ -bisector for $angle ABC$ and $CP$ -bisector for $angle ACB$. Now, it must calculated the area for the triangle $PQR$ knowing that $AB=c$, $BC=a$ and $CA=b$.
I tried to use the angle bisector theorem for every bisectors but I didn't obtained anything.
Thanks :)
geometry
asked Jan 3 '13 at 0:44
IuliIuli
3,66413078
$endgroup$
add a comment |
$begingroup$
How can I calculate the area of a triangle determined by the interior bisectors? What I want to say it is represented in the following picture:
$AQ$ is the bisector of the angle $angle BAC$, $BR$ -bisector for $angle ABC$ and $CP$ -bisector for $angle ACB$. Now, it must calculated the area for the triangle $PQR$ knowing that $AB=c$, $BC=a$ and $CA=b$.
I tried to use the angle bisector theorem for every bisectors but I didn't obtained anything.
Thanks :)
geometry
asked Jan 3 '13 at 0:44
IuliIuli
3,66413078
$endgroup$
add a comment |
$begingroup$
How can I calculate the area of a triangle determined by the interior bisectors? What I want to say it is represented in the following picture:
$AQ$ is the bisector of the angle $angle BAC$, $BR$ -bisector for $angle ABC$ and $CP$ -bisector for $angle ACB$. Now, it must calculated the area for the triangle $PQR$ knowing that $AB=c$, $BC=a$ and $CA=b$.
I tried to use the angle bisector theorem for every bisectors but I didn't obtained anything.
Thanks :)
geometry
asked Jan 3 '13 at 0:44
IuliIuli
3,66413078
$endgroup$
How can I calculate the area of a triangle determined by the interior bisectors? What I want to say it is represented in the following picture:
$AQ$ is the bisector of the angle $angle BAC$, $BR$ -bisector for $angle ABC$ and $CP$ -bisector for $angle ACB$. Now, it must calculated the area for the triangle $PQR$ knowing that $AB=c$, $BC=a$ and $CA=b$.
I tried to use the angle bisector theorem for every bisectors but I didn't obtained anything.
Thanks :)
geometry
geometry
asked Jan 3 '13 at 0:44
IuliIuli
3,66413078
asked Jan 3 '13 at 0:44
IuliIuli
3,66413078
asked Jan 3 '13 at 0:44
IuliIuli
3,66413078
asked Jan 3 '13 at 0:44
IuliIuli
3,66413078
asked Jan 3 '13 at 0:44
IuliIuli
3,66413078
3,66413078
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We'll derive the equation using the fact:
$$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, quad (I)$$
Using the angle bisector theorem we get:
$$BP=frac{ac}{a+b},quad (1)$$
$$BR=frac{ac}{b+c}, quad (2)$$
$$CR=frac{ab}{b+c},quad (3)$$
$$CQ=frac{ab}{a+c},quad (4)$$
$$AQ=frac{bc}{a+c},quad (5)$$
and
$$AP=frac{bc}{a+b}. quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=frac{1}{2}absingamma, quad (7)$$
$$A_{PBR}=frac{1}{2}BPcdot BRsinbeta, quad (8)$$
$$A_{RCQ}=frac{1}{2}CRcdot CQsingamma, quad (9)$$
and
$$A_{QAP}=frac{1}{2}AQcdot APsinalpha. quad (10)$$
Let $R$ be the circumradius, we know that:
$$sin alpha = frac{a}{2R}, quad (11)$$
$$sin beta = frac{b}{2R}, quad (12)$$
$$sin gamma = frac{c}{2R}, quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get:
$$A_{PQR}=frac{1}{2} cdot frac{abc}{2R}-frac{1}{2} frac{a^2c^2b}{(a+b)(b+c)2R}-frac{1}{2} cdot frac{a^2b^2c}{(b+c)(a+c)2R}-frac{1}{2} cdot frac{b^2c^2a}{(a+b)(a+c)2R}, Rightarrow$$
$$A_{PQR}=frac{abc}{4R}[1-frac{ac}{(a+b)(b+c)}-frac{ab}{(b+c)(a+c)}-frac{bc}{(a+b)(a+c)}], Rightarrow$$
$$A_{PQR}=frac{abc}{2R}[frac{abc}{(a+b)(b+c)(a+c)}], Rightarrow$$
$$A_{PQR}=A_{ABC}[frac{2abc}{(a+b)(b+c)(a+c)}]$$
Using Heron's formula we are done.
answered Jan 3 '13 at 13:57
RicardoCruzRicardoCruz
3,0131017
$endgroup$
add a comment |
$begingroup$
This triangle has area $$frac{2abc}{(a+b)(a+c)(b+c)}cdot A,$$ where $A$ is the area of the reference triangle with sides $a,b,c$. It
may be called the "Cevian triangle" with respect to the incenter $I$ of the given reference triangle with sides $a,b,c$, or the "incentral triangle."
edited Nov 5 '18 at 6:01
Hrithik Ravi
34
answered Jan 3 '13 at 2:17
coffeemathcoffeemath
27.3k22342
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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$begingroup$
We'll derive the equation using the fact:
$$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, quad (I)$$
Using the angle bisector theorem we get:
$$BP=frac{ac}{a+b},quad (1)$$
$$BR=frac{ac}{b+c}, quad (2)$$
$$CR=frac{ab}{b+c},quad (3)$$
$$CQ=frac{ab}{a+c},quad (4)$$
$$AQ=frac{bc}{a+c},quad (5)$$
and
$$AP=frac{bc}{a+b}. quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=frac{1}{2}absingamma, quad (7)$$
$$A_{PBR}=frac{1}{2}BPcdot BRsinbeta, quad (8)$$
$$A_{RCQ}=frac{1}{2}CRcdot CQsingamma, quad (9)$$
and
$$A_{QAP}=frac{1}{2}AQcdot APsinalpha. quad (10)$$
Let $R$ be the circumradius, we know that:
$$sin alpha = frac{a}{2R}, quad (11)$$
$$sin beta = frac{b}{2R}, quad (12)$$
$$sin gamma = frac{c}{2R}, quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get:
$$A_{PQR}=frac{1}{2} cdot frac{abc}{2R}-frac{1}{2} frac{a^2c^2b}{(a+b)(b+c)2R}-frac{1}{2} cdot frac{a^2b^2c}{(b+c)(a+c)2R}-frac{1}{2} cdot frac{b^2c^2a}{(a+b)(a+c)2R}, Rightarrow$$
$$A_{PQR}=frac{abc}{4R}[1-frac{ac}{(a+b)(b+c)}-frac{ab}{(b+c)(a+c)}-frac{bc}{(a+b)(a+c)}], Rightarrow$$
$$A_{PQR}=frac{abc}{2R}[frac{abc}{(a+b)(b+c)(a+c)}], Rightarrow$$
$$A_{PQR}=A_{ABC}[frac{2abc}{(a+b)(b+c)(a+c)}]$$
Using Heron's formula we are done.
answered Jan 3 '13 at 13:57
RicardoCruzRicardoCruz
3,0131017
$endgroup$
add a comment |
$begingroup$
We'll derive the equation using the fact:
$$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, quad (I)$$
Using the angle bisector theorem we get:
$$BP=frac{ac}{a+b},quad (1)$$
$$BR=frac{ac}{b+c}, quad (2)$$
$$CR=frac{ab}{b+c},quad (3)$$
$$CQ=frac{ab}{a+c},quad (4)$$
$$AQ=frac{bc}{a+c},quad (5)$$
and
$$AP=frac{bc}{a+b}. quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=frac{1}{2}absingamma, quad (7)$$
$$A_{PBR}=frac{1}{2}BPcdot BRsinbeta, quad (8)$$
$$A_{RCQ}=frac{1}{2}CRcdot CQsingamma, quad (9)$$
and
$$A_{QAP}=frac{1}{2}AQcdot APsinalpha. quad (10)$$
Let $R$ be the circumradius, we know that:
$$sin alpha = frac{a}{2R}, quad (11)$$
$$sin beta = frac{b}{2R}, quad (12)$$
$$sin gamma = frac{c}{2R}, quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get:
$$A_{PQR}=frac{1}{2} cdot frac{abc}{2R}-frac{1}{2} frac{a^2c^2b}{(a+b)(b+c)2R}-frac{1}{2} cdot frac{a^2b^2c}{(b+c)(a+c)2R}-frac{1}{2} cdot frac{b^2c^2a}{(a+b)(a+c)2R}, Rightarrow$$
$$A_{PQR}=frac{abc}{4R}[1-frac{ac}{(a+b)(b+c)}-frac{ab}{(b+c)(a+c)}-frac{bc}{(a+b)(a+c)}], Rightarrow$$
$$A_{PQR}=frac{abc}{2R}[frac{abc}{(a+b)(b+c)(a+c)}], Rightarrow$$
$$A_{PQR}=A_{ABC}[frac{2abc}{(a+b)(b+c)(a+c)}]$$
Using Heron's formula we are done.
answered Jan 3 '13 at 13:57
RicardoCruzRicardoCruz
3,0131017
$endgroup$
add a comment |
$begingroup$
We'll derive the equation using the fact:
$$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, quad (I)$$
Using the angle bisector theorem we get:
$$BP=frac{ac}{a+b},quad (1)$$
$$BR=frac{ac}{b+c}, quad (2)$$
$$CR=frac{ab}{b+c},quad (3)$$
$$CQ=frac{ab}{a+c},quad (4)$$
$$AQ=frac{bc}{a+c},quad (5)$$
and
$$AP=frac{bc}{a+b}. quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=frac{1}{2}absingamma, quad (7)$$
$$A_{PBR}=frac{1}{2}BPcdot BRsinbeta, quad (8)$$
$$A_{RCQ}=frac{1}{2}CRcdot CQsingamma, quad (9)$$
and
$$A_{QAP}=frac{1}{2}AQcdot APsinalpha. quad (10)$$
Let $R$ be the circumradius, we know that:
$$sin alpha = frac{a}{2R}, quad (11)$$
$$sin beta = frac{b}{2R}, quad (12)$$
$$sin gamma = frac{c}{2R}, quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get:
$$A_{PQR}=frac{1}{2} cdot frac{abc}{2R}-frac{1}{2} frac{a^2c^2b}{(a+b)(b+c)2R}-frac{1}{2} cdot frac{a^2b^2c}{(b+c)(a+c)2R}-frac{1}{2} cdot frac{b^2c^2a}{(a+b)(a+c)2R}, Rightarrow$$
$$A_{PQR}=frac{abc}{4R}[1-frac{ac}{(a+b)(b+c)}-frac{ab}{(b+c)(a+c)}-frac{bc}{(a+b)(a+c)}], Rightarrow$$
$$A_{PQR}=frac{abc}{2R}[frac{abc}{(a+b)(b+c)(a+c)}], Rightarrow$$
$$A_{PQR}=A_{ABC}[frac{2abc}{(a+b)(b+c)(a+c)}]$$
Using Heron's formula we are done.
answered Jan 3 '13 at 13:57
RicardoCruzRicardoCruz
3,0131017
$endgroup$
We'll derive the equation using the fact:
$$A_{PQR}=A_{ABC}-A_{PBR}-A_{RCQ}-A_{QAP}, quad (I)$$
Using the angle bisector theorem we get:
$$BP=frac{ac}{a+b},quad (1)$$
$$BR=frac{ac}{b+c}, quad (2)$$
$$CR=frac{ab}{b+c},quad (3)$$
$$CQ=frac{ab}{a+c},quad (4)$$
$$AQ=frac{bc}{a+c},quad (5)$$
and
$$AP=frac{bc}{a+b}. quad (6)$$
Each mentioned area can be calculated using:
$$A_{PQR}=frac{1}{2}absingamma, quad (7)$$
$$A_{PBR}=frac{1}{2}BPcdot BRsinbeta, quad (8)$$
$$A_{RCQ}=frac{1}{2}CRcdot CQsingamma, quad (9)$$
and
$$A_{QAP}=frac{1}{2}AQcdot APsinalpha. quad (10)$$
Let $R$ be the circumradius, we know that:
$$sin alpha = frac{a}{2R}, quad (11)$$
$$sin beta = frac{b}{2R}, quad (12)$$
$$sin gamma = frac{c}{2R}, quad (13)$$
Now if we substitute all the 13 equations in equation $(I)$ we get:
$$A_{PQR}=frac{1}{2} cdot frac{abc}{2R}-frac{1}{2} frac{a^2c^2b}{(a+b)(b+c)2R}-frac{1}{2} cdot frac{a^2b^2c}{(b+c)(a+c)2R}-frac{1}{2} cdot frac{b^2c^2a}{(a+b)(a+c)2R}, Rightarrow$$
$$A_{PQR}=frac{abc}{4R}[1-frac{ac}{(a+b)(b+c)}-frac{ab}{(b+c)(a+c)}-frac{bc}{(a+b)(a+c)}], Rightarrow$$
$$A_{PQR}=frac{abc}{2R}[frac{abc}{(a+b)(b+c)(a+c)}], Rightarrow$$
$$A_{PQR}=A_{ABC}[frac{2abc}{(a+b)(b+c)(a+c)}]$$
Using Heron's formula we are done.
answered Jan 3 '13 at 13:57
RicardoCruzRicardoCruz
3,0131017
edited Jan 3 '13 at 14:04
answered Jan 3 '13 at 13:57
RicardoCruzRicardoCruz
3,0131017
answered Jan 3 '13 at 13:57
RicardoCruzRicardoCruz
3,0131017
answered Jan 3 '13 at 13:57
RicardoCruzRicardoCruz
3,0131017
3,0131017
add a comment |
add a comment |
$begingroup$
This triangle has area $$frac{2abc}{(a+b)(a+c)(b+c)}cdot A,$$ where $A$ is the area of the reference triangle with sides $a,b,c$. It
may be called the "Cevian triangle" with respect to the incenter $I$ of the given reference triangle with sides $a,b,c$, or the "incentral triangle."
edited Nov 5 '18 at 6:01
Hrithik Ravi
34
answered Jan 3 '13 at 2:17
coffeemathcoffeemath
27.3k22342
$endgroup$
add a comment |
$begingroup$
This triangle has area $$frac{2abc}{(a+b)(a+c)(b+c)}cdot A,$$ where $A$ is the area of the reference triangle with sides $a,b,c$. It
may be called the "Cevian triangle" with respect to the incenter $I$ of the given reference triangle with sides $a,b,c$, or the "incentral triangle."
edited Nov 5 '18 at 6:01
Hrithik Ravi
34
answered Jan 3 '13 at 2:17
coffeemathcoffeemath
27.3k22342
$endgroup$
add a comment |
$begingroup$
This triangle has area $$frac{2abc}{(a+b)(a+c)(b+c)}cdot A,$$ where $A$ is the area of the reference triangle with sides $a,b,c$. It
may be called the "Cevian triangle" with respect to the incenter $I$ of the given reference triangle with sides $a,b,c$, or the "incentral triangle."
edited Nov 5 '18 at 6:01
Hrithik Ravi
34
answered Jan 3 '13 at 2:17
coffeemathcoffeemath
27.3k22342
$endgroup$
This triangle has area $$frac{2abc}{(a+b)(a+c)(b+c)}cdot A,$$ where $A$ is the area of the reference triangle with sides $a,b,c$. It
may be called the "Cevian triangle" with respect to the incenter $I$ of the given reference triangle with sides $a,b,c$, or the "incentral triangle."
edited Nov 5 '18 at 6:01
Hrithik Ravi
34
answered Jan 3 '13 at 2:17
coffeemathcoffeemath
27.3k22342
edited Nov 5 '18 at 6:01
Hrithik Ravi
34
edited Nov 5 '18 at 6:01
Hrithik Ravi
34
edited Nov 5 '18 at 6:01
Hrithik Ravi
34
34
answered Jan 3 '13 at 2:17
coffeemathcoffeemath
27.3k22342
answered Jan 3 '13 at 2:17
coffeemathcoffeemath
27.3k22342
answered Jan 3 '13 at 2:17
coffeemathcoffeemath
27.3k22342
27.3k22342
add a comment |
add a comment |
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