Circles tangent to BC at B and C

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Circles tangent to BC at B and C

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#1 h Jun 22, 2024, 3:46 PM • 2 Y

Y by Rounak_iitr, GeoKing

Let $ABC$ be a triangle, and let $\omega_1,\omega_2$ be centered at $O_1$ , $O_2$ and tangent to line $BC$ at $B$ , $C$ respectively. Let line $AB$ intersect $\omega_1$ again at $X$ and let line $AC$ intersect $\omega_2$ again at $Y$ . If $Q$ is the other intersection of the circumcircles of triangles $ABC$ and $AXY$ , then prove that lines $AQ$ , $BC$ , and $O_1O_2$ either concur or are all parallel.

Advaith Avadhanam

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#2 h Jun 22, 2024, 3:47 PM • 1 Y

Y by GeoKing

Solution

This post has been edited 1 time. Last edited by MarkBcc168, Jun 22, 2024, 3:47 PM

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DottedCaculator

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DottedCaculator

#3 h Jun 22, 2024, 11:10 PM • 1 Y

Y by GeoKing

By Monge on $\omega_1$ , $\omega_2$ , and the circle passing through $A$ tangent to the line through $A$ parallel to $BC$ with radius $0$ , we need to show $\frac{BQ}{QC}=\frac{BX}{CY}$ , which follows from spiral similarity as $\triangle QBX\sim\triangle QCY$ .

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#4 h Jun 23, 2024, 4:43 AM • 1 Y

Y by GeoKing

Let $b=-1$ , $c=1$ , $x=-1+s$ , and $y=1+t$ .
We have $\frac{a+1}s,\frac{a-1}t\in\mathbb R$ , so $\frac{\overline sa+\overline s-s}s=\overline a=\frac{\overline ta-\overline t+t}t$ . Solving, $a=\frac{2st-s\overline t-\overline st}{\overline st-s\overline t}$ .
We have $q=\frac{cx-by}{c+x-b-y}=\frac{s+t}{s-t}$ .
Also, letting $q_1=o_1+1$ and $q_2=o_2-1$ , we have $(q_1-s)(\overline{q_1}-\overline s)=q_1\overline{q_1}$ ; substituting $\overline{q_1}=-q_1$ and solving gives $q_1=\frac{s\overline s}{\overline s-s}$ .
The exsimilicenter, which is the intersection of $O_1O_2$ and $BC$ , is given by
$z=\frac{o_1\overline{o_2}-\overline{o_1}o_2}{o_1-o_2-\overline{o_1}+\overline{o_2}}=\frac{(q_1-1)(-q_2+1)-(-q_1-1)(q_2+1)}{(q_1-1)-(q_2+1)-(-q_1-1)+(-q_2+1)}=\frac{2(q_1+q_2)}{2(q_1-q_2)}=\frac{s\overline s(\overline t-t)+t\overline t(\overline s-s)}{s\overline s(\overline t-t)-t\overline t(\overline s-s)}$ .
We are done because this equals the intersection of $AQ$ and $BC$ ,
$\frac{a\overline q-\overline aq}{a-q-\overline a+\overline q}=\frac{\frac{2st-s\overline t-\overline st}{\overline st-s\overline t}\cdot\frac{\overline s+\overline t}{\overline s-\overline t}-\frac{-2\overline{st}+s\overline t+\overline st}{\overline st-s\overline t}\cdot\frac{s+t}{s-t}}{\frac{2st-s\overline t-\overline st}{\overline st-s\overline t}-\frac{s+t}{s-t}-\frac{-2\overline{st}+s\overline t+\overline st}{\overline st-s\overline t}+\frac{\overline s+\overline t}{\overline s-\overline t}}=\frac{(s-t)(2st-s\overline t-\overline st)(\overline s+\overline t)-(\overline s-\overline t)(-2\overline{st}+s\overline t+\overline st)(s+t)}{(s-t)(\overline s-\overline t)((2st-2s\overline t-2\overline st+2\overline{st}))+(\overline st-s\overline t)((s-t)(\overline s+\overline t)(\overline s-\overline t)(s+t))}=\frac{2(s-t-\overline s+\overline t)(s\overline s(\overline t-t)+t\overline t(\overline s-s))}{2(s-t-\overline s+\overline t)(s\overline s(\overline t-t)-t\overline t(\overline s-s))}=z$

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CyclicISLscelesTrapezoid

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CyclicISLscelesTrapezoid

#5 h Jul 9, 2024, 6:50 AM • 1 Y

Y by GeoKing

Lemma: Let $A$ , $B$ , $C$ , and $D$ be points on a circle, and let $\overline{AB}$ and $\overline{CD}$ intersect at $X$ . Then, $\tfrac{CX}{XD}=\tfrac{CA}{AD} \cdot \tfrac{CB}{BD}$ .

Proof: We have
$\frac{CX}{XD}=\frac{[ABC]}{[ABD]}=\frac{\frac{1}{2} CA \cdot CB \cdot \sin \angle ACB}{\frac{1}{2} DA \cdot DB \cdot \sin \angle ADB}=\frac{CA}{AD} \cdot \frac{CB}{BD},$ as desired. $\square$

Let $\overline{AQ}$ and $\overline{BC}$ intersect at $Z$ . Notice that $QBX \sim QCY$ , so $\tfrac{QB}{QC}=\tfrac{BX}{CY}$ . By the lemma, we have
$\frac{ZB}{ZC}=\frac{BA}{AC} \cdot \frac{BQ}{QC}=\frac{AB}{AC} \cdot \frac{BX}{CY}=\frac{AB}{AC} \cdot \frac{2BO_1 \cdot \cos \angle O_1BA}{2CO_2 \cdot \cos \angle O_2CA}=\frac{AB}{AC} \cdot \frac{BO_1}{CO_2} \cdot \frac{\sin \angle ABC}{\sin \angle ACB}=\frac{BO_1}{CO_2}.$ Thus, we have $ZBO_1 \sim ZCO_2$ , so $Z$ , $O_1$ , and $O_2$ are collinear, as desired. $\square$

Remark: This problem has 3 main components: the circles and intersections, the definition of $Q$ , and the definition of $Z$ . These can be dealt with mostly independently of each other: using the lemma, length relations involving $Z$ decompose to those involving $Q$ , which decompose by spiral similarity to those involving $X$ and $Y$ , which are easily calculated.

This post has been edited 1 time. Last edited by CyclicISLscelesTrapezoid, Jul 9, 2024, 6:23 PM

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#6 h Jul 9, 2024, 6:20 PM • 1 Y

Y by GeoKing

Here's a solution using degree 1 moving points / spiral similarity.

It is well-known that spiral similarities are linear maps. In fact, the following lemma is true.

Lemma. (well-known) Let $l_1,l_2$ denote two lines intersecting in point $Z$ . Let $O$ be a fixed point. If $X_t$ is a linearly moving point on $l_1$ and $Y_t=l_2\cap (AOX_t)$ , then $\frac{X_0X_t}{Y_0Y_t}$ is constant.
Proof. $O$ is the center of the spiral similarity sending $X_t$ to $Y_t$ . Hence, $\triangle QX_0X_t\sim QY_0Y_t$ , which means $\frac{X_0X_t}{Y_0Y_t}=\frac{QX_0}{QY_0}$ , which is indeed a constant. $\blacksquare$

Fix point $Q$ on $(ABC)$ . Choose a point $X$ line $AB$ and define $Y=(AQX)\cap AC$ . Then, $X\rightarrow Y$ is a linear map (spiral similarity centered at $Q$ ). From our lemma, we may write $\frac{BX}{CY}=\frac{QB}{QC}$ .
Since $\omega_1,\omega_2$ are tangent to $BC$ , we have $\measuredangle BO_1X=2\measuredangle CBX$ and $\measuredangle CO_2Y=2\measuredangle BCY$ . As a result, we have
$\frac{O_1B}{O_2C}=\frac{BX}{CY}\cdot\frac{\sin(\measuredangle CBA)}{\sin(\measuredangle ABC)}=\frac{QB}{QC}\cdot\frac{AC}{AB}$ Let $T=AQ\cap BC$ . For $T,O_1,O_2$ to be collinear, it suffices to show that $\frac{TB}{TC}=\frac{O_1B}{O_2C}=\frac{QB}{QC}\cdot\frac{AC}{AB},$ which is a well-known identity (sketch: use sine rule in $\triangle TAB,\triangle TAC$ ). $\blacksquare$

This post has been edited 2 times. Last edited by Pyramix, Jul 9, 2024, 6:21 PM

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#7 h Jul 10, 2024, 8:43 AM

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another way to obtain ratio

This post has been edited 1 time. Last edited by Deadline, Jul 10, 2024, 8:45 AM
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mcmp

#8 h Nov 1, 2024, 8:54 AM

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Oh bruh why am I so washed at geometry again :censored:

:censored:

my solution is wayyy overcomplicated oops.

So we start by defining $T=\overline{AQ}\cap\overline{BC}$ . Note $|\operatorname{Pow}_\omega(T)|=TA\cdot TQ=TB\cdot TC=k^2$ , so we can consider an inversion $I$ centred at $T$ with radius $\sqrt{k}$ . Let $Y’=I(X)$ and $X’=I(Y)$ under this inversion. Note that $T-X-Y’$ and $T-X’-Y$ . However I would like to show that $Y’\in\omega_2$ and $X’\in\omega_1$ for reasons to be disclosed later on. This is not hard; notice that $\omega_1\iff\omega_2$ under $I$ which is clear since they are both tangent to $\overline{BC}$ and $B\iff C$ .

Now I can disclose why I wanted $X’$ and $Y’$ to satisfy these properties; it’s quite clear now that under $I$ we must have $\omega_1\iff\omega_2$ , so $\overline{O_1O_2}$ also passes through $T$ .

Edit: I just realised I proved wayyyy too much again (I proved that five lines concurred at $T$ instead of your normal three). To prove your normal three just take the inversion $I$ centred at $T$ with power $\operatorname{Pow}_\omega(T)$ and then note that $\omega_1\iff\omega_2$ , done.

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#9 h Feb 13, 2025, 6:23 PM • 2 Y

Y by Taha.kh, Radin.AmirAslani

Beautiful problem!
Lemma : In a circle $w$ , we have two lines such that one of them intersect $w$ at points $C$ and $B$ and the other intersect $w$ at points $X$ and $Y$ . That two lines intersect at $Z$ ( $Z$ is out of the circle) . Suppose that $B$ lies between $Z$ and $C$ and $X$ lies between $Z$ and $Y$ .
I claim that $\frac{XB}{XC} . \frac{YB}{YC} = \frac{ZB}{ZC}$ ( It follows from similarity)

Now call $T$ the intersection point of AQ and BC . Now applying the lemma :
$\frac{AB}{AC} . \frac{QB}{QC} = \frac{TB}{TC}$ Denote the radius of circle tangent to $BC$ at $B$ R1 and the other one R2
So if we show $\frac{TB}{TC} = \frac{R1}{R2}$ the problem will solve by homothety and that three lines are concur.
By similarity , $\frac{QB}{QC} = \frac{BX}{CY}$ We know that $\frac{a}{sin\angle A} = 2R$ So $\frac{2R1}{2R2} = \frac{BX}{CY} . \frac{sin\angle C}{sin\angle B}$
And the RHS is equivalent to $\frac{QB}{QC} . \frac{AB}{AC}$ . So the problem solved

Note that if R1 = R2 that lines will be parallel and we can check it easily .

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#10 h Mar 30, 2025, 1:20 AM

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wow advaith orz

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