rhombus ABCD construction given circumradii of ABC, BCD

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rhombus ABCD construction given circumradii of ABC, BCD

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Source: 2004 Austria Beginners' Competition p4

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#1 h Oct 4, 2022, 10:10 AM • 3 Y

Y by Mango247, Mango247, Mango247

Of a rhombus $ABCD$ we know the circumradius $R$ of $\Delta ABC$ and $r$ of $\Delta BCD$ . Construct the rhombus.

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#2 h Nov 11, 2022, 8:48 PM • 3 Y

Y by Mango247, Mango247, Mango247

If given the rhombus $ABCD\ :\ A(0,a),B(-b,0),C(0,-a),D(b,0)$ , then $\frac{R}{r}=\frac{a}{b}, a=\frac{2R^{2}r}{R^{2}+r^{2}}$ and $b=\frac{2Rr^{2}}{R^{2}+r^{2}}$ .
$AD=\sqrt{a^{2}+b^{2}}=\frac{2Rr}{\sqrt{R^{2}+r^{2}}}$ .

Construction:
Points $O(0,0),K(r,0),L(0,R)$ , see drawing.
The circle, midpoint $K$ and radius $R$ , intersects the circle, midpoint $L$ and radius $r$ , in the points $A$ and $D$ .
$AD$ is the side of the wanted rhombus.
The circle, midpoint $A$ and radius $AD$ intersects the little circle in the point $B$ .

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PROF65

#3 h Nov 12, 2022, 7:08 PM

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Let $O,I$ the centers of the circles of radius $R,r$
we have $OI \perp BC$ since $I$ is on $AC$ , the perpendicular bisector of $DB$ and $O$ is on $BD$ , the perpendicular bisector of $AC$ then we get $I$ is the orthocenter of $OBC$ if $J$ is the reflecttion of $I$ in $BC$ then $OBCJ$ are cyclic more the center of the circle $(OBC)$ is on $OJ\equiv OJ$ because $OBC,JBC$ are isoceles , whence $OC,CJ$ are perpendicular therefore we can begin the construction :
1) draw the circle $(O)$ and the point $C$ on it
2) draw a segment $CJ$ perpendicular to $OC$ and the circle $(OCJ)$
3) the point $B$ is the second intersection of $(O),(OCJ)$ then draw the line $BO$ and the perpendicular to $BO$ through $C$ we get $A$ as intersection with $(O)$ finally the symmetric of $B$ in $AC$ is $D$

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This post has been edited 1 time. Last edited by PROF65, Nov 12, 2022, 7:15 PM
Reason: rectification of the diagram

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