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Itoz

#1 h Apr 20, 2025, 2:00 PM

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Given a fixed circle $\omega$ with its center $O$ . There are two fixed points $B, C$ and one moving point $A$ on $\omega$ . The midpoint of the line segment $BC$ is $M$ . $R$ is a fixed point on $\omega$ . Line $AO$ intersects $\odot(AMR)$ at $P(\ne A)$ , and line $BP$ intersects $\odot(BOC)$ at $Q(\ne B)$ .

Find all the fixed points $R$ such that $\omega$ is always tangent to $\odot (OPQ)$ when $A$ varies.
Hint

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This post has been edited 2 times. Last edited by Itoz, Apr 20, 2025, 2:02 PM

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#2 h Apr 20, 2025, 10:59 PM • 1 Y

Y by Pengu14

Complex bash with $\omega$ as the unit circle, so that
$|a|=|b|=|c|=|r|=1$ $o = 0$ $m = \frac{b+c}2$ Let $U$ be the circumcenter of $\triangle AMR$ , so that
$u = \frac{ar(m\overline{m}-1)}{m-a-r+ar\overline{m}} = \frac{ar(b-c)^2}{2(abr+acr-2abc-2bcr+b^2c+bc^2)}$ and
$\begin{align*} p &= 2\left[\frac{a + (-a) + u - a(-a)\overline{u}}2\right] - a \\ &= u + a^2\overline{u} - a \\ &= \frac{ar(b-c)^2}{2(abr+acr-2abc-2bcr+b^2c+bc^2)} + \frac{a^2(b-c)^2}{2(abr+acr-2abc-2bcr+b^2c+bc^2)} - a \\ &= \frac{ar(b-c)^2 + a^2(b-c)^2 - 2a(abr+acr-2abc-2bcr+b^2c+bc^2)}{2(abr+acr-2abc-2bcr+b^2c+bc^2)} \\ &= \frac{a(ab^2+2abc+ac^2-2abr-2acr-2b^2c-2bc^2+b^2r+2bcr+c^2r)}{2(abr+acr-2abc-2bcr+b^2c+bc^2)} \\ &= \frac{a(b+c)(ab+ac-2ar-2bc+br+cr)}{2(abr+acr-2abc-2bcr+b^2c+bc^2)} \end{align*}$ Then we have
$p-b = \frac{a^2b^2+2a^2bc+a^2c^2-2a^2br-2a^2cr+2ab^2c-2abc^2-ab^2r+ac^2r+4b^2cr-2b^3c-2b^2c^2}{2(abr+acr-2abc-2bcr+b^2c+bc^2)}$ and so, letting line $BP$ intersect $\omega$ again at $T$ , we have
$t = \frac{b-p}{b\overline{p}-1} = -\frac{p-b}{b(\overline{p}-\overline{b})} = \frac{a(a^2b^2+2a^2bc+a^2c^2-2a^2br-2a^2cr+2ab^2c-2abc^2-ab^2r+ac^2r+4b^2cr-2b^3c-2b^2c^2)}{2a^2br+2a^2cr-4a^2bc-ab^3+abc^2+2ab^2r-2abcr+2b^3c+2b^2c^2-b^3r-2b^2cr-bc^2r}$ Also let $V$ be the circumcenter of $\triangle BOC$ , so that
$v = \frac{bc}{b+c}$ Then
$\begin{align*} q &= 2\left(\frac{b+t+v-bt\overline{v}}2\right) - b \\ &= t + v - bt\overline{v} \\ &= t + \frac{bc}{b+c} - \frac{bt}{b+c} \\ &= \frac{c(b+t)}{b+c} \\ &= \frac{c(a^3b^2+2a^3bc+a^3c^2-2a^3br-2a^3cr-2a^2b^2c-2a^2bc^2+a^2b^2r+2a^2bcr+a^2c^2r-ab^4+2ab^2cr-2ab^3c+2ab^3r-ab^2c^2+2b^4c+2b^3c^2-b^4r-2b^3cr-b^2c^2r)}{(b+c)(2a^2br+2a^2cr-4a^2bc-ab^3+abc^2+2ab^2r-2abcr+2b^3c+2b^2c^2-b^3r-2b^2cr-bc^2r)} \\ &= \frac{c(a^3b+a^3c-2a^3r-2a^2bc+a^2br+a^2cr-ab^3-ab^2c+2ab^2r+2b^3c-b^3r-b^2cr)}{2a^2br+2a^2cr-4a^2bc-ab^3+abc^2+2ab^2r-2abcr+2b^3c+2b^2c^2-b^3r-2b^2cr-bc^2r} \\ &= \frac{c(a+b)(a-b)(ab+ac-2ar-2bc+br+cr)}{2a^2br+2a^2cr-4a^2bc-ab^3+abc^2+2ab^2r-2abcr+2b^3c+2b^2c^2-b^3r-2b^2cr-bc^2r} \end{align*}$ Now let $X$ denote the circumcenter of $\triangle OPQ$ , so that
$x = \frac{pq(\overline{p}-\overline{q})}{\overline{p}q-p\overline{q}}$ Now we compute
$\begin{align*} \overline{p} &= \frac{(b+c)(ab+ac-2ar-2bc+br+cr)}{2a(abr+acr-2abc-2bcr+b^2c+bc^2)} = \frac{p}{a^2} \\ \overline{q} &= \frac{(a+b)(a-b)(ab+ac-2ar-2bc+br+cr)}{a(a^2b^2+2a^2bc+a^2c^2-2a^2br-2a^2cr+2ab^2c-2abc^2-ab^2r+ac^2r+4b^2cr-2b^3c-2b^2c^2)} = \frac{q}{ct} \end{align*}$ Then
$x = \frac{pq\left(\frac{p}{a^2}-\frac{q}{ct}\right)}{\frac{pq}{a^2}-\frac{pq}{ct}} = \frac{a^2q-ctp}{a^2-ct}$ Now
$\begin{align*} a^2q - ctp &= \frac{a^2c(a+b)(a-b)(ab+ac-2ar-2bc+br+cr)}{2a^2br+2a^2cr-4a^2bc-ab^3+abc^2+2ab^2r-2abcr+2b^3c+2b^2c^2-b^3r-2b^2cr-bc^2r} - \frac{a^2c(b+c)(ab+ac-2ar-2bc+br+cr)(a^2b^2+2a^2bc+a^2c^2-2a^2br-2a^2cr+2ab^2c-2abc^2-ab^2r+ac^2r+4b^2cr-2b^3c-2b^2c^2)}{2(abr+acr-2abc-2bcr+b^2c+bc^2)(2a^2br+2a^2cr-4a^2bc-ab^3+abc^2+2ab^2r-2abcr+2b^3c+2b^2c^2-b^3r-2b^2cr-bc^2r)} \\ &= \frac{a^2c(ab+ac-2ar-2bc+br+cr)\left[2(a^2-b^2)(abr+acr-2abc-2bcr+b^2c+bc^2) - (b+c)(a^2b^2+2a^2bc+a^2c^2-2a^2br-2a^2cr+2ab^2c-2abc^2-ab^2r+ac^2r+4b^2cr-2b^3c-2b^2c^2)\right]}{2(abr+acr-2abc-2bcr+b^2c+bc^2)(2a^2br+2a^2cr-4a^2bc-ab^3+abc^2+2ab^2r-2abcr+2b^3c+2b^2c^2-b^3r-2b^2cr-bc^2r)} \\ &= \frac{a^2c(ab+ac-2ar-2bc+br+cr)(2a^3br+2a^3cr-4a^3bc+2a^2b^2r+2a^2c^2r-a^2b^3-a^2b^2c-a^2bc^2-a^2c^3-ab^3r-ab^2cr-abc^2r-ac^3r+2ab^3c+2abc^3-4b^2c^2r+2b^3c^2+2b^2c^3)}{2(abr+acr-2abc-2bcr+b^2c+bc^2)(2a^2br+2a^2cr-4a^2bc-ab^3+abc^2+2ab^2r-2abcr+2b^3c+2b^2c^2-b^3r-2b^2cr-bc^2r)} \end{align*}$ and
$\begin{align*} a^2-ct &= a^2 - \frac{ac(a^2b^2+2a^2bc+a^2c^2-2a^2br-2a^2cr+2ab^2c-2abc^2-ab^2r+ac^2r+4b^2cr-2b^3c-2b^2c^2)}{2a^2br+2a^2cr-4a^2bc-ab^3+abc^2+2ab^2r-2abcr+2b^3c+2b^2c^2-b^3r-2b^2cr-bc^2r} \\ &= \frac{a(2a^3br+2a^3cr-4a^3bc+2a^2b^2r+2a^2c^2r-a^2b^3-a^2b^2c-a^2bc^2-a^2c^3-ab^3r-ab^2cr-abc^2r-ac^3r+2ab^3c+2abc^3-4b^2c^2r+2b^3c^2+2b^2c^3)}{2a^2br+2a^2cr-4a^2bc-ab^3+abc^2+2ab^2r-2abcr+2b^3c+2b^2c^2-b^3r-2b^2cr-bc^2r} \end{align*}$ Thus we have a huge amount of cancellation, and what remains is
$x = \frac{ac(ab+ac-2ar-2bc+br+cr)}{2(abr+acr-2abc-2bcr+b^2c+bc^2)}$ Now since the circumcircle of $\triangle OPQ$ passes through $O$ , it is tangent to $\omega$ if and only if $|x| = \frac12$ , or equivalently if
$|ab+ac-2ar-2bc+br+cr| = |abr+acr-2abc-2bcr+b^2c+bc^2|$ $|b+c-2r|\left|a - \frac{2bc-br-cr}{b+c-2r}\right| = |br+cr-2bc|\left|a - \frac{bc(b+c-2r)}{2bc-br-cr}\right|$ Since $|b+c-2r| = |br+cr-2bc|$ , and since this is nonzero because $r\neq \frac{b+c}2$ because $M$ lies inside $\omega$ , we cancel it out to see that $|x| = \frac12$ if and only if
$\left|a - \frac{2bc-br-cr}{b+c-2r}\right| = \left|a - \frac{bc(b+c-2r)}{2bc-br-cr}\right|$ That is, as $a$ varies on the unit circle, $a$ needs to be equidistant from both of these points. But if the two points are distinct, then their perpendicular bisector intersects the unit circle in at most two points. So the only way for this to work is if those two points are the same point -- that is,
$\frac{2bc-br-cr}{b+c-2r} = \frac{bc(b+c-2r)}{2bc-br-cr}$ $(2bc-br-cr)^2 = bc(b+c-2r)^2$ $4b^2c^2 - 4bcr(b+c) + r^2(b+c)^2 = bc(b+c)^2 - 4bcr(b+c) + 4bcr^2$ $r^2(b-c)^2 = bc(b-c)^2$ $r^2 = bc$ and so $R$ is the midpoint of one of the two arcs $BC$ on $\omega$ . $\blacksquare$

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#3 h Apr 21, 2025, 10:49 AM

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Suppose that $R$ is not on the perpendicular bisector of $BC$ take $A$ as the reflection of $R$ on this line. Let $N$ be the midpoint of arc $BRC$ and let $ML$ be a diameter of $(RAPM)$ . Clearly $L$ is outside $\omega$ .

By power of point

$OP = \frac{OL\cdot OM}{OR}$ let $d$ be the diameter of $(OPQ)$ , by sine law
$d = \frac{OP}{sin(\angle OQP)} = \frac{OP}{sin(\angle OCB)} = \frac{OP}{\frac{OM}{OC}} = \frac{OP\cdot OC}{OM}$
Combinig all we get
$d = \frac{\frac{OL\cdot OM}{OR}\cdot OC}{OM} = OL > ON.$
this means $(OPQ)$ intersects $\omega$ twice. Same computation shows that $R$ being the midpoint of both arcs determined by $BC$ works.

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