Hunt for that Region! 2.0

Realizing that the given figure is a right triangle, we could solve the problem graphically quite easily. First we place the triangle in a cartesian coordinate system as so:

We pick a random point $P = (x, y)$ where $x = P_x$ and $y = P_y$ . From the illustration the mentioned areas could be written as:

$S_{\triangle PAB} = \frac{45y}{2}$

$S_{\triangle PBC} = \frac{60x}{2}$

$S_{\triangle PAC} = S_{\triangle ABC} - S_{\triangle PBC} - S_{\triangle PAB} = \frac{45*60}{2} - \frac{60x}{2} - \frac{45y}{2}$

Rearranging and solving the given inequalities we get:

$90y > 45*60 - 60x \Rightarrow 2x + 3y > 90$ $(1)$

$45*60 - 45y < 120x \Rightarrow 8x + 3y > 180$ $(2)$

$45y > 60x \Rightarrow 3y > 4x$ $(3)$

Since both $(1)$ and $(2)$ are linear inequalities we could easily find their solutions on the coordinate system using the extreme cases. Consequently the region that satisfies $(1)$ is the area $(with$ $\color{#20A900} green)$ above the line $y = -\frac{2}{3} x + 30$ , for $(2)$ it is the area $(with$ $\color{#EC7300} orange)$ under $y = -\frac{8}{3} x + 60$ , for $(3)$ - the $\color{#69047E}purple$ region above line $y = \frac{4x}{3}$ .

It so happens that all three functions intersect at one point D. The region that we are looking for is $\triangle CED$ . We find the coordinates of point D by putting the functions of the aforementioned lines equal to each other :

$-\frac{2}{3} x + 30 = -\frac{8}{3} x + 60 \Rightarrow 2x = 30 \Rightarrow D = (15, 20)$ .

Point E is the intersection between $CA$ and $y = \frac{4x}{3}$ . Therefore the x coordinate of E is the solution to the equation $-\frac{4x}{3} + 60 = \frac{4x}{3} \Rightarrow x_E = 22.5; y_E = 30 \Rightarrow CE = 37.5$ . The distance from point D to $CA$ can be found using the already mentioned formulas for the three areas. It turns out to be equal to 12. Finally, we can calculate the formula using the formula $\frac{CE*12}{2} = 225$

P.S. Looking into the illustration I've provided above, it turns out that the point D is the center of mass of the triangle. This gives insight into a more elegant solution to the problem. Let P be a random point in the triangle and $CP \cap BA = M$ . The ratio between $S_{\triangle PBC}$ and $S_{\triangle PAC}$ is just $\frac{BM}{MA}$ . In order for that to be greater than one (this is what the problem implies) point M has to be in the right half of BA. In other words every point on a line to the right of the median from point C provides us with a solution to one of the inequalities. Doing this for all medians/inequalities, the area in question once again turns out to be $\triangle CED$ . Using ratios $S_{\triangle CED} = \frac{1}{2} * S_{\triangle CDA} = \frac{1}{2} * \frac{2}{3} * S_{\triangle CNA}$ (point N is the midpoint of BA) $= \frac{1}{2} * \frac{2}{3} * \frac{1}{2} * S_{\triangle CBA} = \frac{1}{6} * S_{\triangle CBA} = 225.$

In the diagram above, point $G$ is defined such that area of $\Delta GAB=$ area of $\Delta GBC=$ area of $\Delta GAC$

The desired region is the green triangle $\Delta CGH$ .

Area of $\Delta ABC=\cfrac{45\cdot60}{2}=1350$

area of $\Delta GAB=$ area of $\Delta GBC=$ area of $\Delta GAC=\cfrac{1350}{3}=450$

$\cfrac{1}{2}\cdot GF\cdot45=\cfrac{1}{2}\cdot GE\cdot60=\cfrac{1}{2}\cdot GJ\cdot75=450$

$GF=20$ , $GE=15$ , $GJ=12$

$EBFG$ is a rectangle.

$\therefore EB=GF=20$

$\tan \angle EBG=\cfrac{GE}{EB}=\cfrac{15}{20}=\cfrac{3}{4}$

$\tan \angle BCA=\cfrac{45}{60}=\cfrac{3}{4}$

$\angle GHJ=\angle EBG+\angle BCA$

$\therefore \tan \angle GHJ=\tan (\angle EBG+\angle BCA)=\cfrac{\cfrac{3}{4}+\cfrac{3}{4}}{1-(\cfrac{3}{4})^2}=\cfrac{\cfrac{3}{2}}{\cfrac{7}{16}}=\cfrac{3}{2}\cdot\cfrac{16}{7}=\cfrac{24}{7}$

$\tan \angle GHJ=\cfrac{GJ}{HJ}=\cfrac{24}{7}$

$\cfrac{12}{HJ}=\cfrac{24}{7}$

$HJ=\cfrac{12\cdot7}{24}=\cfrac{7}{2}$

$CE+EB=CB$

$CE=60-20=40$

By $\textbf{Pythagorean Theorem}$ ,

$CE^2+GE^2=CG^2$

$GJ^2+CJ^2=CG^2$

$\therefore CE^2+GE^2=GJ^2+CJ^2$

$40^2+15^2=CJ^2+12^2$

$CJ^2=40^2+(15^2-12^2)=40^2+9^2=41^2$

$CJ=41$

$CJ=CH+HJ$

$CH=41-\cfrac{7}{2}=\cfrac{75}{2}$

$\therefore$ Area of desired region $=$ Area of $\Delta CGH$ $=\cfrac{1}{2}\cdot\cfrac{75}{2}\cdot{12}=225$