area inequality inside a right triangle (1996 Denmark MO - Mohr Contest p1)

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area inequality inside a right triangle (1996 Denmark MO - Mohr Contest p1)

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#1 h Aug 19, 2020, 7:00 AM • 1 Y

Y by Mango247

In triangle $ABC$ , angle $C$ is right and the two catheti are both length $1$ . For one given the choice of the point $P$ on the cathetus $BC$ , the point $Q$ on the hypotenuse and the point $R$ are plotted on the second cathetus so that $PQ$ is parallel to $AC$ and $QR$ is parallel to $BC$ . Thereby the triangle is divided into three parts. Determine the locations of point $P$ for which the rectangular part has a larger area than each of the other two parts.

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#2 h Aug 19, 2020, 4:05 PM

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We think of the instance where quadrilateral $RQPC$ is a square. Obviously $\frac12\cdot\frac12>\frac12\cdot\frac12\cdot\frac12$ .
We then think of the instance where quadrilateral $RQPC$ is a non-square rectangle. Say we increase $CP$ by $x$ . So then we can change this inequality to
$(\frac12+x)(\frac12-x)>(x+\frac12)^2\frac12$ because side $RC$ would decrease by $x$ (you can prove this by coordinates, sum of $QP$ & $CP$ needs to equal one of the side lengths to be on the hypotenuse if slope of hypotenuse $=1$ ). We find the greatest value of x for the rectangle to have a larger area than the larger triangle and we know our range of solutions.
$\Rightarrow 0>(4x-1)(3x+1)$ We want $x\in \mathbb{R}^+$ , so we have $x\in(0,\frac14)$ .
Thus, $P\in(\frac14,\frac34)$ away from $C$ for this to work. $\blacksquare$

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This post has been edited 1 time. Last edited by thedodecagon, Aug 19, 2020, 4:10 PM

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