parallelogram wanted, square, midpoint, perpendicular, angle bisector related

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Source: Lithuania TST 2018 p5

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#1 h Sep 30, 2020, 1:22 PM • 1 Y

Y by Mango247

The midpoint $M$ is marked on the side $AB$ of the square $ABCD$ . From the point $B$ , the perpendicular onthe line $CM$ intersects the line $CM$ at point $P$ . The point $N$ divides segment $CP$ in half. The angle bisector of $DAN$ and the segment $DP$ intersect at point $Q$ . Prove that the quadrilateral $BMQN$ is parallelogram.

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#2 h Sep 30, 2020, 1:57 PM

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Cartesian coordinates solve it easily.

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#3 h Sep 30, 2020, 2:12 PM • 3 Y

Y by Mango247, Mango247, Mango247

Yup nice coordinate bash exercise!

Let $A=(0,10),B=(0,0),C=(10,0),D=(10,10)$ . Then $M=(0,5),P=(2,4),N=(6,2),Q=(6,7)$ . Now it is easy to see that $BMQN$ is obviously a parallelogram.

In case anyone is wondering, you find $P$ by noting that the equation of $MC$ is $y=-\frac 12 +5$ . Then the equation of $BP$ must be $y=2x$ . Solving the system, you easily find the coordinates of $P$ .

To find $Q$ , note that $AN=10$ by $6-8-10$ Pythagorean triples. Then $\triangle DAN$ is isosceles so the angle bisector of $\angle DAN$ is the equation of $AE$ where $E=(8,6)$ , the midpoint of $DN$ . Then find the equation of $AE$ and $DP$ and solve to find $Q$ .

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#4 h Sep 30, 2020, 4:25 PM

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Let $A=(0,0), B=(0,10), C=(10,10) D=(10,0)$ . We can easily find that $MC$ is $y=\frac{x}{2}+5$ and since $BP\perp CM$ , the slope of $BM$ is $-2$ Hence, $BP$ is $-2x+10$ and we can just solve the system $\begin{cases} -2x+10=y\\ \frac{x}{2}+5=y\end{cases}$ and find the coordinates of $P$ , which happen to be $P=(2,6)$ .
$N=(\frac{2+10}{2},\frac{6+10}{2})=(6,8)$ .
Let $Q$ be the midpoint of $DP$ . $NQ$ is the midline of $\triangle CDP$ and so $NQ\parallel CD\parallel BM$ and $NQ=\frac{CD}{2}=\frac{AB}{2}=BM$ . Thus, $BMQN$ is parallelogram. $AN$ is $8x-6y=0$ and $AD$ is $y=0$
Now, by point-line distance formula, we check that $Q=(6,3)$ is equidistant from lines $AD$ and $AN$ and hence lies on the bisector. That is $\frac{|(8)(6)+(-6)(3)|}{\sqrt{8^2+6^2}}=\frac{3}{1}$ which is true.

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#5 h Sep 30, 2020, 5:58 PM • 2 Y

Y by Mango247, Mango247

Weirdly enough, this was also Romanian NMO 2017/3. The author listed on the Romanian papers is Adrian Bud. Here is the solution I gave when working on the problems some time ago.

We first prove the following

Lemma. Let $XYZW$ be a square and $T$ the midpoint of side $XY$ . If $S$ is the projection of $Y$ onto $TZ$ , then $SW=ZW$ .

Proof. We denote $XT=TY=x$ .

It's clear that $YZ=ZW=TW=2x$ .

By the Pythagorean theorem in triangle $TYZ$ we have $TZ^2=TY^2+YZ^2=5x^2\Leftarrow TZ=x\sqrt{5}$ .

On the other hand, it's well known that $TY^2=TS\cdot TZ$ and $YZ^2=SZ\cdot TZ$ , so $\frac{TS}{SZ}=\frac{TY^2}{YZ^2}=\frac{1}{4}\Leftarrow 4TS=SZ$ . Noting that $TS+SZ=TZ=x\sqrt{5}$ we easily compute $TS=\frac{x}{\sqrt{5}}$ as well as $SZ=\frac{4x}{\sqrt{5}}$ .

Applying the Pythagorean theorem in triangle $TWX$ we obtain $WT^2=XT^2+XW^2\Leftarrow WT=x\sqrt{5}$ .

By Stewart's theorem in triangle $WTZ$ for cevian $WS$ we obtain
$WS^2\cdot TZ=WZ^2\cdot TS+WT^2\cdot SZ-TS\cdot SZ\cdot TZ$ $WS^2\cdot x\sqrt{5}=4x^2\cdot\frac{x}{\sqrt{5}}+5x^2\cdot\frac{4x}{x\sqrt{5}}-\frac{x}{\sqrt{5}}\cdot\frac{4x}{\sqrt{5}}\cdot x\sqrt{5}$ $WS^2=4x^2\Rightarrow WS=2x$
Finally $WS=ZW$ , which is just what we wanted to prove.

Back to the main problem.

Let $R$ be the intersection of $CM$ and $AD$ .

It's easily seen that $RACB$ is a parallelogram and therefore $AR=AD$ .

Using the lemma above, we have $DP=DC$ and since $N$ is the midpoint of $PC$ , we have that $DN\perp PC$ .

Now since $AR=AD$ and $\angle RND=90^{\circ}$ it follows that $AN=AD=AR$ .

In particular, since $AN=AD$ and $[AQ$ is the bisector of angle $\angle DAN$ , it follows that $Q$ is on the perpendicular bisector of $DN$ , which gives $QN=QD$ . Since $\angle PND=90^{\circ}$ , we have that $NQ$ is a median in triangle $PND$ so $Q$ is the midpoint of $DP$ .

Now since $Q$ and $N$ are midpoints of segments $DP$ and $PC$ respectively, $QN$ is a midline in $PDC$ and therefore

$\Rightarrow QN\parallel DC$ but since $AB\parallel DC$ we have that $QN\parallel AB\Leftrightarrow QN\parallel MB$

$\Rightarrow QN=\frac{DC}{2}$ but since $DC=AB$ and $MB=\frac{AB}{2}$ we have $QN=MB$ .

The above two claims prove that $BMQN$ is a parallelogram, so we are done.

This post has been edited 3 times. Last edited by trigadd123, Oct 22, 2020, 8:52 AM

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