Expected Area

Notice that Q is distributed uniformly on BC, since the probability that Q lies in interval is proportional to the length of the interval.

Let x denote the distance of Q from B. So the area is equal to $\frac{xh}{2} = x \frac{A}{13} = \frac{84x}{13}$ by Heron's Formula .

Now, this means we need to integrate $\frac{84x}{13*13}$ from 0 to 13, and it is easy to see that this comes out to 42.

A method of obtaining randomly distributed points on this triangle is to make it into a parallelogram by reflecting its image along AC then choosing points uniformly along lines parallel to AB and BC. When such points lie in the reflected image they can be discarded. Viewed in this way it may be seen that the base length of the random triangle BAQ is uniformly distributed on [0,13] and its height is a constant. The random vertical position of P is immaterial. Hence the required area is just half the area of the given triangle.

The expected value of $[ABQ]$ is the average value of all the possible areas for $ABC$ . However, consider this. For each area $[ABQ]$ that is possible, there is a corresponding possible $[ABQ']$ which they sum to the area of the triangle. Hence, for all possible triangle areas, there is a corresponding area that will sum to $[ABC]$ . This implies that the average is half the area of the triangle.

To solve for the area of the triangle, we can use Heron's formula: $[ABC] = \sqrt{s(s-a)(s-b)(s-c)}$ , $s$ is half the sum of the side lengths.

$[ABC] = \sqrt{(21)(6)(7)(8)}$

$[ABC] = 84$

$[ABQ_{expected}] = \frac {84}{2}$

$[ABQ_{expected}] = 42$

Heron's formula says that the area of a triangle with sides $a, b,$ and $c$ is $\sqrt{(s)(s-a)(s-b)(s-c)}$ , where $s=\frac{a+b+c}{2}$ , or the semiperimeter. The semiperimeter of $ABC$ is $\frac{13+14+15}{2}=\frac{42}{2}=21$ . Therefore, the area is $\sqrt{(21)(21-13)(21-14)(21-15)}=\sqrt{21 \cdot 8 \cdot 7 \cdot 6}=$ $\sqrt{3 \cdot 7 \cdot 2 \cdot 2 \cdot 2 \cdot 7 \cdot 2 \cdot 3}=\sqrt{(2^2)(2^2)(3^2)(7^2)}=(2)(2)(3)(7)=84$ .

Unless [ABQ] is half [ABC] (P lies on Q), for any point $P$ extended to $Q$ , there is always a point $P' \neq Q$ extended to $Q' \neq Q$ such that $[ABQ]=84-[ABQ']$ . Therefore, the points "even out", and every pair of these points give us a total area of 84; the expected value of one of these areas is $\frac{84}{2}=\boxed{42}$

By Heron's formula which states that the area if a triangle equals [(s)(s-a)(s-b)(s-c)]^(1/2) where s denotes the semiperimeter of triangle ABC, a,b,c denotes the sidelengths of the triangle, we can easily calculate the area of triangle to be [(21)(8)(7)(6)]^(1/2)=84. Since the area of triangle ABQ is given by the formula (k/2)(BQ), where k denotes the length of perpendicular dropped from A to CB, we find that the area of triangle AQB is directly proportional to the length of BQ, that is if we plot the gtaph of area against length of BQ, we will get a straight line passing through the origin. Since the point Q lands on BC randomly (evident from the fact that point P is chosen randomly), the expected value of triangle ABQ is (1/2)(84)=42.

Since we have to find the expected area of ABQ, we can see that AQ divides the triangle into two parts (not necessarily equal). Therefore the expected value is the same as the mean value of area ie half the area of the triangle ABC. By Heron's Formula we find the area of triangle ABC to be 84 units. Therefore our ans is 42.

We have $[ABQ]+[ACQ]=[ABC]$ for all positions of $P$ . Therefore $E([ABQ])+E([ACQ])=[ABC]$ . On the other hand, by symmetricity, we have $E([ABQ])=E([ACQ])$ . Therefore, $E([ABQ])=E([ACQ])=[ABC]/2=42$ .

Splitting triangle ABC into a 5-12-13 and 9-12-15 triangles, we can easily see that its area is 84. Now consider the median from A. By definition, BC is bisected; this line splits the triangle into two smaller ones of equal area. If these areas are equal, a randomly selected point has equal probability of landing in either one. Averaging the two, the point is expected to land on the median. Therefore, the triangle's base is halved and our answer is 84/2=42.

The area of $\triangle ABC$ is given by Heron's formula, with semiperimeter $s = \frac{1}{2}(a+b+c) = \frac{1}{2}(13+14+15) = 21$ : $|\triangle ABC| = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = 84.$ Next, consider the proportion $\frac{|\triangle ABQ|}{|\triangle ABC|} = \frac{BQ}{BC}$ . But any random point $P$ uniformly chosen in the interior of $\triangle ABC$ is also uniformly distributed on the line $B'C' || BC$ through $P$ , so $\frac{BQ}{BC} = \frac{B'P}{B'C'}$ , so the expected value of this proportion is simply $\frac{1}{2}$ , from which it follows that the expected value of the area of $\triangle ABQ$ is half the area of $\triangle ABC$ , or $42$ .

$$ First, we obtain $\,\angle BAC$ by using cosine's formula. $$ $\cos \theta=\frac{14^2+15^2-13^2}{2\cdot 14\cdot 15}=\frac{3}{5}\;\;\;\Rightarrow\;\;\;\sin \theta=\frac{4}{5}.$ $$ Then, draw $\;\Delta \text{ABC}$ in first quadrant of Cartesian diagram and set $\,A$ as the origin. Therefore, we have $\,A(0,0)$ , $\,B(9,12)$ , and $\,C(14,0)$ . The coordinate of $\,B\,$ is calculated by using length of $\,AB\,$ and $\,\angle BAC$ . $$ If the coordinate of its vertex is known, the area of a triangle $\,[(x_1,y_1)$ , $\;(x_2,y_2)$ , $\;(x_3,y_3)]$ can be obtained by using formula below: $$ $A=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|.$ $$ To calculate area of $\,\Delta \text{ABQ}$ , we have $\,A(0,0)$ , $\,B(9,12)$ , and $\,Q(x,y)$ , where it's very clear that $\,9\le x \le 14$ and $\,0\le y \le 12$ since $\,Q$ is a point that lies on line $\,BC$ . We have $$ $A_{\Delta \text{ABQ}}=\frac{1}{2}|0(12-y)+9(y-0)+x(0-12)|=\frac{1}{2}|9y-12x|.$ $$ By using statistical approach, if we assume point $\;Q(x,y)$ is independently and identically uniformly distributed, the joint probability density function of point $\;Q(x,y)$ is $$ $f(x,y)=f(x)\cdot f(y)=\frac{1}{(14-9)}\cdot \frac{1}{(12-0)}=\frac{1}{60}.$ $$ Thus, the expected value of area $\;\Delta \text{ABQ}$ is $$ $\begin{aligned} \text{E} [A_{\Delta \text{ABQ}}]&=\left|\int\int A(x,y)f(x,y)\,dxdy\right|\\ &=\left|\int_{y=0}^{12}\int_{x=9}^{14} \frac{1}{2}(9y-12x)\cdot \frac{1}{60}\,dxdy\right|\\ &=\frac{1}{120}\left|\int_{y=0}^{12}\left.(9yx-6x^2)\right|_{x=9}^{14}\,dy\right|\\ &=\frac{1}{120}\left|\int_{y=0}^{12}(45y-690)\,dy\right|\\ &=\frac{1}{120}\left|\left.(\frac{45}{2}y^2-690y)\right|_{y=0}^{12}\right|\\ &=\frac{1}{120}\cdot|-5040|\\ &=\boxed{42} \end{aligned}$ $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$