What is the largest possible area that $PQRS$ could have?

Question

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?

Source.

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!

For starters, links to images are strongly discouraged, especially low quality images. I would recommend that you type your posts. Formatting tips here. — Em., Jul 23 '16 at 06:43
Thanks for the tips! I really appreciate the edit. Will keep all this in my mind next time! — allquiet1984, Jul 23 '16 at 09:45
@TEKNO Don't mind the haters, there's nothing wrong with your question. — Gregory Grant, Jul 23 '16 at 12:46
@GregoryGrant Thanks man! Was bad linking on my part though. Any idea about the question? — allquiet1984, Jul 24 '16 at 03:10
@TEKNO It seems to me that it has to be largest when $AP$ and $DS$ are both equal to one, in which case the answer is $222$ — Gregory Grant, Jul 24 '16 at 13:23

score 0 · Answer 1 · edited Jul 28 '23 at 16:04

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!

What is the largest possible area that $PQRS$ could have?

1 Answers1