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problem from Kyiv math festival

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Source: 6-th Kyiv math festival, 2007

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rogue

#1 h May 9, 2007, 1:10 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$

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#2 h May 10, 2007, 4:11 AM • 4 Y

Y by Adventure10, Mango247, and 2 other users

A very nice construct problem !

Quote:

Are given a triangle $ABC$ and a point $D\in (BC)$ . Construct the points $E\in (CA)$ and $F\in (AB)$ so that
the middlepoints of the segments $[DE]$ and $[DF]$ are collinearly with the vertex $C$ and
the middlepoints of the segments $[DE]$ and $[EF]$ are collinearly with the vertex $A$ .

The proper construction.

1. Construct the reflection $X$ of the point $D$ w.r.t. the point $C$ .
2. Construct the point $Y\in AB$ for which $DY\parallel AC$ .
3. Construct the point $E\in CA$ so that $YE\parallel BC$ .
4. Construct the point $F\in XE\cap AB$ .

Remark. The positions of the points $E\in (CA)$ , $F\in (AB)$ are defined by the following relations :
$\boxed{\frac{EA}{EC}=\frac{DC}{DB}}$ and $\boxed{\frac{AF}{AB}=\left(\frac{CD}{CB}\right)^{2}}$ . Denote the reflection $Z$ of the point $Y$ w.r.t. the point $A$ .
The above second relation applied in the trapezoid $DYZX$ , where $AC$ is the middle line, i.e. $\frac{AF}{AB}=\left(\frac{YF}{YB}\right)^{2}$ ,

shows that the point $F$ is the conjugate of the point $B$ w.r.t. the pair $\{Y,Z\}$ , i.e. $FI\parallel AC$ , where $I\in XY\cap DZ$ .

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rogue

#3 h May 11, 2007, 6:15 PM • 2 Y

Y by Adventure10, Mango247

Nice solution, Virgil Nicula!

My solution is even more simple:

Quote:

The point $D$ at the side $AB$ of triangle $ABC$ is given. Construct points $E,F$ at sides $BC, AC$ respectively such that the midpoints of $DE$ and $DF$ are collinear with $B$ and the midpoints of $DE$ and $EF$ are collinear with $C.$

Let $K,L$ and $M$ be the midpoints of $DE, EF$ and $DF$ and let $BM$ intersects $AC$ at point $G.$ Then $BG\parallel EF,$ triangles $BCG$ and $ECF$ are similar and $KC$ is a median of triangle $BCG.$ Therefore $K$ is common midpoint of $DE$ and $BG,$ i.e. $BEGD$ is parallelogram.
Construction: draw $DG\parallel BC,$ then $GE\parallel AB$ and $EF\parallel BG.$
Proof:By the construction the intersection point of $BG$ and $DE$ is the midpoint of $DE.$ Since $EF\parallel BG$ the intersection point of $KC$ and $EF$ is the midpoint of $EF.$ Finally, the intersection point of $BG$ and $DF$ is the midpoint of $DF.$
$\begin{picture}(100,105) \put(10,10){\qbezier{(0,0)(50,0)(100,0)}\qbezier{(0,0)(20,40)(40,80)}\qbezier{(40,80)(70,40)(100,0)}\qbezier{(10,20)(33,40)(55,60)}\qbezier{(44,0)(49,30)(55,60)}\qbezier{(10,20)(27,10)(44,0)}\qbezier{(27,10)(34,45)(40,80)}\qbezier{(32,40)(67,20)(100,0)}\put(0,0){\circle*{2}}\put(100,0){\circle*{2}}\put(40,80){\circle*{2}}\put(10,20){\circle*{2}}\put(55,60){\circle*{2}}\put(44,0){\circle*{2}}\put(33,40){\circle*{2}}\put(49,30){\circle*{2}}\put(27,10){\circle*{2}}\put(25,0){\circle*{2}}\put(-7,-12){A}\put(103,-12){C}\put(37,82){B}\put(1,21){D}\put(56,62){E}\put(50,-12){F}\put(24,41){K}\put(53,30){L}\put(18,16){M}\put(17,-12){G}\linethickness{.1pt}{\qbezier{(25,0)(18,10)(10,20)}\qbezier{(25,0)(40,30)(55,60)}\qbezier{(25,0)(33,40)(40,80)}}}\end{picture}$

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