Construct a rectangle

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Construct a rectangle

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Source: Czech and Slovak MO, 2002, P5

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Sayan

#1 h Sep 15, 2012, 3:44 AM • 2 Y

Y by Adventure10 and 1 other user

A triangle $KLM$ is given in the plane together with a point $A$ lying on the half-line opposite to $KL$ . Construct a rectangle $ABCD$ whose vertices $B, C$ and $D$ lie on the lines $KM, KL$ and $LM$ , respectively. (We allow the rectangle to be a square.)

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yetti

#2 h Oct 15, 2012, 1:24 PM • 2 Y

Y by Adventure10, Mango247

Put coordinate origin at $A$ and positive x-axis along the ray $(AK$ . Let $K = (k, 0)$ and $\kappa = \tan \measuredangle MKL$ $\Longrightarrow$
equation of line $MK$ is $y = \kappa(x-k)$ . Let $B \in MK$ be arbitrary point $\Longrightarrow$ $B = (x_B, y_B)$ , where $y_B = \kappa(x_B-k)$ .
Perpendicular bisector of $[AB]$ cuts $AK$ at center $O$ of circle $(O)$ with radius $[OA] = [OB]$ . $BO$ cuts $(O)$ again at $D = (x_D, y_D)$ , where $y_D = -y_B = -\kappa(x_B-k)$ .
Let $B_0$ be perpendicular projection of $B$ on $AK$ . Power of $B_0$ to $(O)$ is $-x_B \cdot x_D = y_B \cdot y_D = -y_D^2$ .
Equation of locus $\mathcal D$ of $D$ is obtained by eliminating the arbitrary parameter $x_B$ : $\frac{y_D^2}{x_D} = x_B = -\frac{y_D}{\kappa} + k$ $\Longrightarrow$ $h(x,y) = y^2 + \frac{xy}{\kappa} - kx = 0$ .
This is general equation $ax^2 + bxy + cy^2 + dx + ey + f = 0$ of a conic, with $a = e = f = 0$ . Since $b^2 - 4ac = \frac{1}{\kappa^2} > 0$ , it is a hyperbola.
Slopes of its asymptotes $u, v$ are obtained by factoring the first 3 terms $ax^2 + bxy + cy^2 = y^2 + \frac{xy}{\kappa} = y \left(y + \frac{x}{\kappa}\right) = 0$ $\Longrightarrow$
their slopes are $0$ and $-\frac{1}{\kappa}$ $\Longrightarrow$ one asymptote is $u \parallel AK$ , the other is $v \perp MK$ . Let their equations be $y = p, y = -\frac{x}{\kappa} + q$ , resp. For any $(x,y) \in u \cup v$ ,
$0 = (y - p) \left(y+\frac{x}{\kappa} - q\right) = h(x,y) - \left(\frac{p}{\kappa} - k\right)x - (p + q)y + pq$ and $h(x,y) = \text{const}$ $\Longrightarrow$ $h(x,y) = -pq$ and $p = \kappa k$ , $q = -p = -\kappa k$ .
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Let $MK$ cut y-axis at $Q$ and let $P$ be reflection of $Q$ in $A$ . According to the above, asymptote $u \parallel AK$ through $P$ and asymptote $v \perp MK$ through $Q$ .
Since absolute term of the hyperpola equation is $0$ $\Longrightarrow$ $A \in \mathcal D$ . Since $A$ is midpoint of $[PQ]$ $\Longrightarrow$ $PQ$ is tangent of $\mathcal D$ at $A$ .
$X \equiv u \cap v$ is hyperbola center. Internal bisector $w$ of $\measuredangle PXQ$ is hyperbola major axis line. Let $E, F \in w$ be hyperbola foci (to be constructed).
Hyperbola tangent $PAQ$ , bisecting $\measuredangle FAE$ , cuts perpendicular bisector of $[EF]$ through $X$ at $S$ . Circumcircle $(Y)$ of $\triangle AEF$ , centered on $XS$ and passing through $A, S,$
cuts hyperbola major axis line $w$ at hyperbola foci $E, F.$ Perpendicular projection $E_0$ of the focus $E$ on hyperbola tangent $PQ$ lies on hyperbola pedal circle $(X)$ ,
which cuts hyperbola major axis line $w$ at hyperbola vertices $U, V$ .
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Let $(E)$ be circle centered at $E$ with radius $[UV]$ . Let $F'$ be reflection of $F$ in $LM$ . The 2 circles $(D_1), (D_2)$ through $F, F'$ and externally tangent to $(E)$ at $T_1, T_2$ are centered on the line $LM$ .
Since $[D_1E] - [D_1F] = [D_1E] - [D_1T_1] = [T_1E] = [UV]$ and similarly, $[D_2E] - [D_2F] = [UV]$ $\Longrightarrow$ $D_1, D_2 \in \mathcal D$ are intersections of the line $LM$ and hyperbola $\mathcal D$ .
Let $(G)$ be arbitrary circle through $F, F',$ preferably intersecting $(E)$ at $J, J'$ . Radical axes $FF', JJ'$ of circle pairs $(D_1), (G)$ and $(E), (G)$ meet at radical center $R$ of $(D_1), (E), (G)$ .
Tangents $RT_1, RT_2$ of $(E)$ from $R$ are radical axes of externally tangent circle pairs $(D_1), (E)$ and $(D_2), (E)$ $\Longrightarrow$ $D_1 \equiv ET_1 \cap LM$ and $D_2 \equiv ET_2 \cap LM$ .
Perpendicular bisectors of $[AD_1], [AD_2]$ cut the line $AK$ at circumcenters $O_1, O_2$ of rectangles $AB_1C_1D_1, AB_2C_2D_2$ , resp., the 2 problem solutions.

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