Distance Tells Us The Area

Let $E$ be a point on $AB$ where $PE$ is perpendicular to $AB$ . Let $AB$ be distance $a$ , $PE$ be distance $x$ , and $EB$ be distance $y$ . From the problem we have $x^{2}+y^{2}=23^{2}$ and similarly, $(a-x)^{2}+(a-y)^{2}=29^{2}$ .

$(a-x)^{2}+(a-y)^{2}=29^{2}\Rightarrow 2a^{2}-2ax-2ay+x^{2}+y^{2}=841$

Substitute $x^{2}+y^{2}=23^{2}$ to get:

$2a^{2}-2ax-2ay+529=841\Rightarrow a^{2}-ax-ay=156$

$a^{2}$ is double the area of $\Delta ABC$ , $ax$ is double the area of $\Delta ABP$ , and $ay$ is double the area of $\Delta BPC$ . Recognize that the area of $\Delta APC$ = $\Delta ABC - \Delta ABP - \Delta BPC$ . To get this, divide the above equation by 2. The answer is thus $\frac{1}{2}(a^{2}-ax-ay)=78$ .

We can solve a more general case of the problem using coordinate geometry. At first it seems that how it could be possible to find the area when we are just given the distance $PA$ and $PB$ . We won't even need to evaluate anything else. By using determinant and distance formula we can figure out how the required area relates to the given distances.

So let the coordinates of $A,B,C,D,P$ be $(0,0), (a,0), (a,a), (0,a), (x,y)$ . Let $PB=m$ and $PD=n$ .

Using determinant we find that $[APC]$ $=$ $\frac{1}{2}$ $|a(x-y)|$

Using distance formula we find that, $m^2 = (x-a)^2 + y^2...(1)$ and $n^2 = x^2 + (y-a)^2....(2)$

Now subtracting (2) from (1) yields that $m^2 - n^2 = 2ay - 2ax$ thus $\frac{1}{2}$ $|a(x-y)|$ = $\frac{1}{4}$ $|m^2 - n^2|$

Therefore, $[APC]$ $=$ $\frac{1}{4}$ $|23^2 - 29^2|$ $=$ $78$

Simple extreme cases The smallest square that can accommodate this will have the shorter segment (23) aligned with its side as shown. $29^2 = (23 + x)^2 + x^2$ giving $x = \frac{\sqrt{1153} - 23}{2}$ . The desired area of the yellow triangle is same as the blue triangle (common base, identical height) = $\frac{x (23 + x)}{2} = \frac {1}{2} \frac{\sqrt{1153} - 23}{2} \times \frac{46 + \sqrt{1153} - 23}{2} = \frac{1153 - 23^2}{8} = 78$

Now if we want to be adventurous and want to do away with the condition that P is inside the square, we get an even more interesting special case, where the segments (29) and (23) overlap to give an isosceles triangle of base 6 and height 26 as shown

The invariance of the area can be proved using conics - Ellipse and Hyperbola. Consider a family of ellipses sharing a common major axis 29 + 23 = 52, with varying foci $( \pm f,0 )$ . Coordinates (h,k) of points at a fixed distance (say 23) from one focus, on each of these ellipses has a property $4hf = 29^2 - 23^2 = 312 = constant$ Here (h) is the height of the triangle and 2f the base of the triangle or the diagonal of the square.

Further, the points will be the locus of intersections of the family of ellipses mentioned above and a family of hyperbolae (with fixed vertices as a = constant = 3 = (29 - 23)/2) I will try to animate it and post a video. @Calvin Lin what is the best way to post this? Not everyone sees a solution.

The cheat way of solving this problem is to assume that DPB can be a straight line, with a length of 23 + 29 = 52. The center of the diagonal is 3 from point P, so the area of the yellow triangle is (1/2)(3)(52) = 78

See Hasan's proof showing the area of the yellow triangle is invariant. Thank you, Hasan.

Let h be the altitude of triangle CPB. h is drawn from P perpendicularly to CB, so the area of |CPB| is (1/2)hs, where s is the side of the square. Similarly, let m be the altitude from P to AB, so the |APB| = (1/2)ms.By the Pythagorean Theorem: (1) (s - h)^2 + (s - m)^2 =29^2 = 841 = s^2 - 2sh + h^2 + s^2 - 2sm + m^2 = 841, and (2) h^2 + m^2 = 23^2 = 529. Substituting (2) into (1), and simplifying, s^2 - s(h + m) = 156, or s(h + m) = s^2 - 15. The area of |APC| = (1/2)s^2 - |APB| - |CPB| = (1/2)s^2 - (1/2)ms - (1/2)hs, or (s/2)[s - (m + h)]; but m + h = s - 156/s. So |APC| =(s/2)*(156/s) = 78. Ed Gray