In a $14\times 18$ rectangle $ABCD$, points $P,Q,R$ and $S$ are chosen, one on each side $ABCD$ as pictured. The lengths $AP, PB, BQ, QC, CR, RD, DS$ and $SA$ are all positive integers and $PQRS$ is a rectangle. What is the largest possible area that $PQRS$ could have?
I am aware that you can guess and check to solve the question but this is under examination conditions and so theoretically I would want to be able to solve this in under 5 minutes in order to consider this solved in the test. I was wondering if any of you had an inkling of a logical procedural way to solve this question and if so be able to explain it to me! Whoever replies is greatly appreciated! Thanks!
forgive my formatting and such, i don't usually do these.
You can use the coordinate system to solve this.
We have: AP = RC, PB = DR, SD = BQ, AS = QC
Since we have to make sure PQRS is a triangle, we cannot only use guess and check. It would take too long. The product of the slopes of perpendicular lines = -1, so we can use this instead.
here's the picture of the solution since I have less than 10 reputation right now lol.
Basically, we get slope PS * slope SR = -1 $$\frac{14-y}{18-x}*\frac{y}{-x}=-1$$
Simplify $$14y-y^2=18x-x^2$$ $$y^2-14y=x^2-18x$$ Then, we can try values of y (or x, whichever you prefer. I used y here.) if you try y = 1, x would not have an integer value. This goes on until y=5, where you get an integer value for x. $$5^2-14(5)=x^2-18x$$ $$-45=x^2-18x$$ $$x^2-18x+45=0$$ $$(x-15)(x-3)=0$$ Now, there are 2 values of x (and y). The other value of y is just 14-y, and the other value of x is just 18-x, so it doesn't really matter much. $$(x,y)=(15,5)$$ You can now find the largest possible area for rectangle PQRS. $$14*18-2 (\frac{15*5}2-\frac{3*9}2) = 150$$
the answer is 150, yippee!