Taniya Melton

2022-10-13

Expected area of a triangle where 1 point is within another triangle

Suppose we have triangle ABC with area k with a point P chosen inside ABC. What is the expected area of triangle PBC?

I'm pretty sure if we let P be the centroid we get k/3. Also, how would you solve this question for an n-sided polygon?

Suppose we have triangle ABC with area k with a point P chosen inside ABC. What is the expected area of triangle PBC?

I'm pretty sure if we let P be the centroid we get k/3. Also, how would you solve this question for an n-sided polygon?

Marlene Welch

Beginner2022-10-14Added 23 answers

Step 1

Let the vertices of the triangle be a, b, c. Then the map

$\begin{array}{}\text{(1)}& \mathbf{g}:\phantom{\rule{1em}{0ex}}(u,v)\mapsto \{\begin{array}{rl}x(u,v)& ={a}_{1}+u({b}_{1}-{a}_{1})+v({c}_{1}-{a}_{1})\\ y(u,v)& ={a}_{2}+u({b}_{2}-{a}_{2})+v({c}_{2}-{a}_{2})\text{},\end{array}\end{array}$

maps the arbitrary triangle from

$(x,y)=({a}_{1},{a}_{2}),({b}_{1},{b}_{2}),({c}_{1},{c}_{2})$ bijectively onto a right triangle with $(u,v)=(0,0),(1,0),(0,1)$ (make a sketch for visualization). Therefore the transformation formula for multiple integrals can be used, and we obtain

${\int}_{T}f(x,y)\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}(x,y)={\int}_{{T}^{\prime}}f{\textstyle (}x(u,v),y(u,v){\textstyle )}\phantom{\rule{mediummathspace}{0ex}}{\textstyle |}{J}_{\mathbf{g}}(u,v){\textstyle |}\text{}\mathrm{d}(u,v)\text{}.$

The Jacobian determinant is a constant: From (1) we obtain

${J}_{\mathbf{g}}(u,v)=det\left[\begin{array}{cc}{x}_{u}& {x}_{v}\\ {y}_{u}& {y}_{v}\end{array}\right]=({b}_{1}-{a}_{1})({c}_{2}-{a}_{2})-({c}_{1}-{a}_{1})({b}_{2}-{a}_{2})\text{}.$

Step 2

Therefore we can write

${\int}_{T}f(x,y)\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}(x,y)={\textstyle |}{J}_{\mathbf{g}}{\textstyle |}\phantom{\rule{mediummathspace}{0ex}}{\int}_{0}^{1}\phantom{\rule{mediummathspace}{0ex}}{\int}_{0}^{1-u}f{\textstyle (}x(u,v),y(u,v){\textstyle )}dv\text{}du\text{}.$

The area of an arbitrary triangle a,b,p is $f=1/2\cdot \text{base}\cdot \text{height}$. In u,v coordinates the base is 1 and the height is v. For normalization the integral has to be divided by the total area that is 1/2. The expected area of a random triangle a,b,p is therefore

$\begin{array}{}\text{(2)}& \mathbb{E}[{A}_{\mathbf{a},\mathbf{b},\mathbf{p}}]=\frac{\frac{1}{2}}{\frac{1}{2}}{\textstyle |}{J}_{\mathbf{g}}|{\int}_{0}^{1}{\int}_{0}^{1-u}vdv\text{}du\text{}=\frac{1}{6}{\textstyle |}({b}_{1}-{a}_{1})({c}_{2}-{a}_{2})-({c}_{1}-{a}_{1})({b}_{2}-{a}_{2}){\textstyle |}\end{array}$

As the area of the triangle a, b, c is

${A}_{\mathbf{a},\mathbf{b},\mathbf{c}}=\frac{1}{2}{\textstyle |}({b}_{1}-{a}_{1})({c}_{2}-{a}_{2})-({c}_{1}-{a}_{1})({b}_{2}-{a}_{2}){\textstyle |}$

the expected area of the random triangle is 1/3 of the area of the original triangle, independently which side is used as a base.

Let the vertices of the triangle be a, b, c. Then the map

$\begin{array}{}\text{(1)}& \mathbf{g}:\phantom{\rule{1em}{0ex}}(u,v)\mapsto \{\begin{array}{rl}x(u,v)& ={a}_{1}+u({b}_{1}-{a}_{1})+v({c}_{1}-{a}_{1})\\ y(u,v)& ={a}_{2}+u({b}_{2}-{a}_{2})+v({c}_{2}-{a}_{2})\text{},\end{array}\end{array}$

maps the arbitrary triangle from

$(x,y)=({a}_{1},{a}_{2}),({b}_{1},{b}_{2}),({c}_{1},{c}_{2})$ bijectively onto a right triangle with $(u,v)=(0,0),(1,0),(0,1)$ (make a sketch for visualization). Therefore the transformation formula for multiple integrals can be used, and we obtain

${\int}_{T}f(x,y)\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}(x,y)={\int}_{{T}^{\prime}}f{\textstyle (}x(u,v),y(u,v){\textstyle )}\phantom{\rule{mediummathspace}{0ex}}{\textstyle |}{J}_{\mathbf{g}}(u,v){\textstyle |}\text{}\mathrm{d}(u,v)\text{}.$

The Jacobian determinant is a constant: From (1) we obtain

${J}_{\mathbf{g}}(u,v)=det\left[\begin{array}{cc}{x}_{u}& {x}_{v}\\ {y}_{u}& {y}_{v}\end{array}\right]=({b}_{1}-{a}_{1})({c}_{2}-{a}_{2})-({c}_{1}-{a}_{1})({b}_{2}-{a}_{2})\text{}.$

Step 2

Therefore we can write

${\int}_{T}f(x,y)\phantom{\rule{mediummathspace}{0ex}}\mathrm{d}(x,y)={\textstyle |}{J}_{\mathbf{g}}{\textstyle |}\phantom{\rule{mediummathspace}{0ex}}{\int}_{0}^{1}\phantom{\rule{mediummathspace}{0ex}}{\int}_{0}^{1-u}f{\textstyle (}x(u,v),y(u,v){\textstyle )}dv\text{}du\text{}.$

The area of an arbitrary triangle a,b,p is $f=1/2\cdot \text{base}\cdot \text{height}$. In u,v coordinates the base is 1 and the height is v. For normalization the integral has to be divided by the total area that is 1/2. The expected area of a random triangle a,b,p is therefore

$\begin{array}{}\text{(2)}& \mathbb{E}[{A}_{\mathbf{a},\mathbf{b},\mathbf{p}}]=\frac{\frac{1}{2}}{\frac{1}{2}}{\textstyle |}{J}_{\mathbf{g}}|{\int}_{0}^{1}{\int}_{0}^{1-u}vdv\text{}du\text{}=\frac{1}{6}{\textstyle |}({b}_{1}-{a}_{1})({c}_{2}-{a}_{2})-({c}_{1}-{a}_{1})({b}_{2}-{a}_{2}){\textstyle |}\end{array}$

As the area of the triangle a, b, c is

${A}_{\mathbf{a},\mathbf{b},\mathbf{c}}=\frac{1}{2}{\textstyle |}({b}_{1}-{a}_{1})({c}_{2}-{a}_{2})-({c}_{1}-{a}_{1})({b}_{2}-{a}_{2}){\textstyle |}$

the expected area of the random triangle is 1/3 of the area of the original triangle, independently which side is used as a base.

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