Restore a bicentral quadrilateral if two opposite vertices and the incenter are given.
Problem
Source: Sharygin 2024 Correspondence Round P8
Tags: geometry, bicentric quadrilateral
InterLoop
06.03.2024 14:15
sus compass ruler given
Call the opposite vertices $B$ and $D$, the incenter $I$ and the wanted inradius $r$. Let $1$ be a fixed unit length.
Let $B_1$ and $B_2$ be the touchpoints of the tangents to the wanted incircle from $B$, and $D_1$ and $D_2$ are defined similarly. Note that we want
$$\angle B_1BB_2 = 180^{\circ} - \angle D_1DD_2$$Alternatively, we want
$$\angle IBB_2 = 90^{\circ} - \angle IDD_2$$It is clear that $\sin(\angle IBB_2) = r/IB$ and $\sin(\angle IDD_2) = r/ID$. Thus what we want is
$$\sin(\angle IBB_2) = \cos(\angle IDD_2)$$$$\sin^2(\angle IBB_2) = 1 - \sin^2(\angle IDD_2)$$$$\frac{r^2}{IB^2} = 1 - \frac{r^2}{ID^2}$$$$\frac{1}{IB^2} + \frac{1}{ID^2} = \frac{1}{r^2}$$Claim 1. Given a segment of length $n$, we can construct a segment of length $1/n$ (with ruler and compass)
$\textsf{\textbf{Proof of Claim.}}$ {Draw a right angled triangle (perpendiculars can be drawn with ruler and compass) with the base $n$ and the height $1$. From the vertex which is contained by the base and the hypotenuse, mark a length $1$ on the base, and from that marked point, construct a perpendicular onto the base, extending till the hypotenuse. The length of this segment must be $1/n$ by similar triangles.
$\blacksquare$
Now since we have the lengths $IB$ and $ID$, we can easily construct the lengths $1/IB$ and $1/ID$, and clearly we can also construct a right triangle with the base and height being those lengths. Since
$$\frac{1}{IB^2} + \frac{1}{ID^2} = \frac{1}{r^2}$$The hypotenuse of that triangle must be $1/r$. Taking the reciprocal of this length, we have a segment of length $r$.
Now we construct a circle of radius $r$ around center $I$, draw tangents from $B$ and $D$ to this circle, and let the intersections of the tangents be $A$ and $C$. The quadrilateral $ABCD$ thus obtained is a bicentral quadrilateral.
idkk
06.03.2024 15:12
I guess problem number is wrong , it should be P7 Anyways here is my solution Problem:- Restore a bicentral quadrilateral if two opposite vertices and the incenter are given Say the opposing the vertices are $A,C$ incentre is $I$. Case 1: $C$ does not lie on $AC$ Lemma 1 : Given any bicentral quadrilateral $ABCD$ with incentre $I$ not lying on $AC$ . and if $B$ lies on same side of $AC$ as $I$.,$\angle{ABC}=\angle{AIC}-90$,. This is just angle chasing $\angle{BAI}=\frac{\angle{DAB}}{2}$ similar computation for $BCI$ shows they add to $\frac{180}{2}=90$[opposite angles of cylic are supplementary]. Also as $\angle{AIC}=\angle{BAI}+\angle{BCI}+\angle{ABC}=90+\angle{ABC}$ [just using sum of angles of a quadrilateral=360] so $\angle{ABC}=\angle{AIC}-90$ Say the opposite vertices are $A,C$ and incentre is given as $I$. We first construct the vertex of $ABCD$ lying on same side of $AC$as $I$[say $B$] Lemma 2 : Given a line segment $AB$ a point $C$ not on $AB$ we can construct the reflection $C$ about $AB$. Construct the circle The circle centered at $A$ with radii $AC$ and same for $B$ with radii $BC$ say they meet again at $C'$. Its then easy to see $AC'B$ and $ACB$ are congruent by $SSS$ critetion. Join $CC'$ also then $\angle{C'AB}=\angle{CAB}$ and $CA=C'A$. So $CC'$ is perpendicular on $AB$. and if it intersects $AB$ at $T$ then $TC=TC'$ hence $C'$ is reflection of $C$ about $AB$. Lemma 3:Given any line segment $AB$ we can construct the the unique point $C \neq A$ such that $BA=BC$. Construct the circle with centre $B$ radii $BA$ let $AB$ intersect it again at $A '$ hence $BA=BA'$ giving us the desired point. Lemma 4: Given any line segment $AB$ we can construct its perpendicular bisector. Draw the circle centered at $A$ radii $AB$ same for $B$ let them intersect at $T,T'$ As $BT=AT$ and $BT'=AT'$ and $AB=AB$ So we have $\Delta T'AT $ and $\Delta T'BT$ are congruent. So $TT'$ bisect $\Delta AT'B$ So $TT'$ is perpendicular bisector so just join it. Lemma 5: Given any line segment $AB$ we can construct the line perpendicular on it from $A$. Use lemma 3 to construct $B'$ with $AB'=AB$ and then use lemma 4 to construct perpendicular bisector of $BB'$. Lemma 6: Given any triangle $ABC$ we can construct its circumcentre Construct perpendicular bisectors of $AB$ and $AC$ let them intersect at $O$ then $O$ becomes the desired point. Lemma 7: Given any triangle $ABC$ if the bisector of $\angle{BAC}$ meets the circumcircle of $ABC$ at $M \neq A$ then $M$ lies on the perpendicular bisector of $BC$ By incenter excenter lemma $MC=MB$ hence it must lie on perpendicular bisector of $BC$. With all those lemmas we start progressing on orginal problem. To construct the required vertex[the one lying on same side] say its $B$. We first reflect $I$ about $AC$ say $I'$ now construct the perpendicular on $I'C$ from $C$ say it meets $AI'$ at $T$. Reflect $T$ about $AC$ to get $T'$ Construct the circumcircle of $T'AC$. say its $\omega$ Then as $B$ lies on same side of $AC$as $T'$ and $\angle{AT'C}=\angle{ATC}=\angle{AI'C}-\angle{I'CT}=\angle{AI'C}-90=\angle{AIC}-90=\angle{ABC}$ Hence lie on a circle. We just construct the perpendicular bisector of $AC$ let it intersect that arc thats opposite to $I$ at $M$ and $MI$ meet $\omega$ again at $P_1$ then $P_1$ has to be equal $B$. [by lemma 7] To construct the other vertex namely $D$ We construct the the circumcircle of $ABC$. Construct the perpendicular bisector of $AC$ to meet the arc of the circumcircle of $ABC$ which contains $B$ say it does at $M_2$ as $D$ lies on the cirumcircle of $ABC$ and $DI$ bisects $\angle{ADC}$ . So by lemma 7 $DI$ passes through $M_2$ so to construct it we can just produce $M_2I$ to meet the circumcircle again at some point and which has to be $D$. Case 2: $I$ lies on $AC$. This would imply $CB=CD$ and $AB=AD$ by lemma 7. Taking $AC=AC$ we have a pair of congruent triangles . [SSS congruence] $\Delta ACB ,\Delta ACD$ are ccongruent So $\angle{ABC}=\angle{ADC}$ as its cyclic they sum to $180$ so each of them is $90$. Construct perpendicular bisector of $AC$ let it intersect $AC$ at $O$ Now a draw a circle with center $O$ and $OA$ radius Hence $B,D$ lie on it perpendicular bisector of $AC$ say it intersects the circle at $M_1$ and $M_2$ as $BI$ bisects $\angle{ABC}$ and $DI$ does to $\angle{ADC}$ by producing $M_2I$ and $M_1I$ to meet the circle at $P_1,P_2$ should basically be $B,D$ in some order. however order dosent matter this should give us the required quadrilateral.
Om245
06.03.2024 17:25
Let $P$ be center of $\triangle BID$. Let $O$ is intersection of perpendiculars bisector of $BD$ and $(PBD)$ Claim : $O$ is circumcenter of $\Box ABCD$ $2\measuredangle {BID} =\measuredangle {BPD}$ Note from easy angle chasing we get $\measuredangle {BID}=\measuredangle{BAD}$ hence $\measuredangle {BOD}=2\measuredangle {BAD}=\measuredangle {BPD}$ Draw circle $\omega$ center at $O$ with radius $OB$. Now let $A',C'$ be intersections of $OP$ with $\omega$. Then by well know property $C=C'I\cap\omega$ and $A=A'I\cap\omega$
Professor33
15.01.2025 13:01
InterLoop wrote: sus compass ruler given
Call the opposite vertices $B$ and $D$, the incenter $I$ and the wanted inradius $r$. Let $1$ be a fixed unit length.
Let $B_1$ and $B_2$ be the touchpoints of the tangents to the wanted incircle from $B$, and $D_1$ and $D_2$ are defined similarly. Note that we want
$$\angle B_1BB_2 = 180^{\circ} - \angle D_1DD_2$$Alternatively, we want
$$\angle IBB_2 = 90^{\circ} - \angle IDD_2$$It is clear that $\sin(\angle IBB_2) = r/IB$ and $\sin(\angle IDD_2) = r/ID$. Thus what we want is
$$\sin(\angle IBB_2) = \cos(\angle IDD_2)$$$$\sin^2(\angle IBB_2) = 1 - \sin^2(\angle IDD_2)$$$$\frac{r^2}{IB^2} = 1 - \frac{r^2}{ID^2}$$$$\frac{1}{IB^2} + \frac{1}{ID^2} = \frac{1}{r^2}$$Claim 1. Given a segment of length $n$, we can construct a segment of length $1/n$ (with ruler and compass)
$\textsf{\textbf{Proof of Claim.}}$ {Draw a right angled triangle (perpendiculars can be drawn with ruler and compass) with the base $n$ and the height $1$. From the vertex which is contained by the base and the hypotenuse, mark a length $1$ on the base, and from that marked point, construct a perpendicular onto the base, extending till the hypotenuse. The length of this segment must be $1/n$ by similar triangles.
$\blacksquare$
Now since we have the lengths $IB$ and $ID$, we can easily construct the lengths $1/IB$ and $1/ID$, and clearly we can also construct a right triangle with the base and height being those lengths. Since
$$\frac{1}{IB^2} + \frac{1}{ID^2} = \frac{1}{r^2}$$The hypotenuse of that triangle must be $1/r$. Taking the reciprocal of this length, we have a segment of length $r$.
Now we construct a circle of radius $r$ around center $I$, draw tangents from $B$ and $D$ to this circle, and let the intersections of the tangents be $A$ and $C$. The quadrilateral $ABCD$ thus obtained is a bicentral quadrilateral.
AMAZING SOLUTION BROTHER