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Source: ISL 2018 G7

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Neothehero

18 posts

Neothehero

#1 h Jul 17, 2019, 12:20 PM • 2 Y

Y by centslordm, Adventure10

Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $ABC$ . Let $P$ be an arbitrary point on $\Omega$ , distinct from $A$ , $B$ , $C$ , and their antipodes in $\Omega$ . Denote the circumcentres of the triangles $AOP$ , $BOP$ , and $COP$ by $O_A$ , $O_B$ , and $O_C$ , respectively. The lines $\ell_A$ , $\ell_B$ , $\ell_C$ perpendicular to $BC$ , $CA$ , and $AB$ pass through $O_A$ , $O_B$ , and $O_C$ , respectively. Prove that the circumcircle of triangle formed by $\ell_A$ , $\ell_B$ , and $\ell_C$ is tangent to the line $OP$ .

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Ankoganit

3070 posts

Ankoganit

#2 h Jul 17, 2019, 12:26 PM • 12 Y

Y by Supercali, AlastorMoody, Pluto1708, TheMathsBoy, gabrupro, centslordm, megarnie, PRMOisTheHardestExam, Adventure10, Mango247, Funcshun840, Lasier

This was also India TST day 4 P3. Here's a solution by Ojas Mittal that uses pure angel chase:
Solution

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TheDarkPrince

3042 posts

TheDarkPrince

#3 h Jul 17, 2019, 12:32 PM • 3 Y

Y by starchan, centslordm, Adventure10

My solution in the TST: complex bash (which was extremely clean)

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psi241

49 posts

psi241

#4 h Jul 17, 2019, 12:33 PM • 4 Y

Y by k12byda5h, centslordm, guptaamitu1, Adventure10

Let $S_A,S_B,S_C$ be a intersection of lines tangent to $\Omega$ at $A,B,C$ with a line tangents to $\Omega$ at $P$ . Then clearly $O_A$ is a midpoint of $OS_A$ . Similarly, $O_B$ is midpoint of $OS_B$ and $O_C$ is midpoint of $OS_C$ .
Taking homothety $\mathcal{H}(O,2)$ , lines $\ell_A,\ell_B,\ell_C$ map to lines pass though $S_A,S_B,S_C$ and perpendicular to $BC,AC,AB$ . Let $X_A,X_B,X_C$ be intersections of these lines so that $X_AX_B\perp AB$ , $X_BX_C\perp BC$ and $X_CX_A\perp CA$ . It's enough to prove that $X_AX_BX_C$ tangents to $OP$ . By angle chasing, $X_AX_BX_C\sim ABC$ . Reflect $O$ across $P$ to $O'$ .

Next we claim that $S_AS_BX_CO'$ concyclic. We can prove this by angle chasing
$\measuredangle S_AX_BS_C=-\measuredangle S_CX_BX_C=-\measuredangle ABC =-\measuredangle S_AOS_C=\measuredangle S_AO'S_C.$ Similarly, $S_AS_BX_CO'$ , $X_AS_BS_CO'$ are concyclic. By Miquel's theorem $X_AX_BX_CO'$ is concyclic. Let P' be antipode of P wrt $\Omega$ . By angle chasing again. $\measuredangle O'X_BX_A=\measuredangle O'S_AS_C=-\measuredangle OS_AS_C=90^{\circ}-\measuredangle S_AOP=90^{\circ}-\measuredangle AP'P=\measuredangle P'PA=\measuredangle P'BA.$ So, $X_AX_BX_CO'\sim ABCP'$ . Let $X$ be a circumcenter $X_AX_BX_C$ . Because $\measuredangle(XO',OP')=\measuredangle(AB,X_AX_B)=90^{\circ}$ .
Therefore, circumcircle of $X_AX_BX_C$ tangents to $OP$ at $P'$ .

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G81928128

13 posts

G81928128

#5 h Jul 17, 2019, 12:35 PM • 4 Y

Y by centslordm, megarnie, Adventure10, sabkx

Seriously, this is a G7?

15min solution

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Reason: brackets

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v_Enhance

6877 posts

v_Enhance

#6 h Jul 17, 2019, 12:45 PM • 11 Y

Y by FadingMoonlight, e_plus_pi, JustKeepRunning, MeowX2, v4913, centslordm, HamstPan38825, TheHimMan, Adventure10, Mango247, Funcshun840

In fact, the circumcircle is tangent to line $OP$ at $P$ .

We present two solutions. In both solutions, we let $\ell_A$ , $\ell_B$ , $\ell_C$ determine triangle $XYZ$ .

First solution with complex numbers (Evan Chen) We proceed with complex numbers with $A$ , $B$ , $C$ on the unit circle (hence $O = 0$ ), and assume that the point $P$ has coordinate $p = 1 \in {\mathbb C}$ . We will show that line $OP$ is tangent at $P$ to the circumcircle of $\triangle PYZ$ .

To solve for $Z$ , recall that the tangents to $\Omega$ at $A$ and $P$ meet at $\frac{2ap}{a+p} = \frac{2a}{a+1}$ . Hence, $o_A = \frac{a}{a+1}$ . Since $\overline{ZO_A} \perp \overline{BC}$ we must have $\frac{z - \frac{a}{a+1}}{b-c} = - \overline{\left( \frac{z-\frac{a}{a+1}}{b-c} \right)} = -\frac{\overline z - \frac{1}{a+1}}{\frac{c-b}{bc}} \iff \frac1b \left( z - \frac{a}{a+1} \right) = c\left( \overline z - \frac{1}{a+1} \right).$ Similarly, since $\overline{ZO_B} \perp \overline{AC}$ , $\frac1a \left( z - \frac{b}{b+1} \right) = c\left( \overline z - \frac{1}{b+1} \right)$ . Subtracting the two gives $\begin{align*} \left(\frac 1b - \frac 1a\right) z &= c\left( \frac{1}{b+1}-\frac{1}{a+1} \right) + \frac{a}{b(a+1)} - \frac{b}{a(b+1)} \\ &= \frac{c(a-b)}{(a+1)(b+1)} = \frac{a^2(b+1) - b^2(a+1)}{ab(a+1)(b+1)} \\ &= \frac{abc(a-b) + (a-b)(ab+a+b)}{ab(a+1)(b+1)} = \frac{(a-b)}{ab} \cdot \frac{ab+a+b+abc}{(a+1)(b+1)} \\ \implies z &= \frac{ab+a+b+abc}{(a+1)(b+1)} \implies z-1 = \frac{abc-1}{(a+1)(b+1)}. \intertext{Similarly,} y-1 &= \frac{abc-1}{(a+1)(c+1)}. \end{align*}$ Then take the complex number $\begin{align*} \theta = \frac{y-p}{p-0} \div \frac{y-z}{p-z} &= \frac{(y-1)(z-1)}{z-y} = \frac{\frac{(abc-1)^2}{(a+1)^2(b+1)(c+1)}} {\frac{abc-1}{a+1}\left( \frac{1}{b+1}-\frac{1}{c+1} \right)} = \frac{abc-1}{(a+1)(c-b)}. \intertext{Its conjugate is} \overline\theta &= \frac{\frac{1}{abc}-1}{(\frac1a+1)(\frac1c-\frac1b)} = \frac{1-abc}{(1+a)(b-c)} = \theta. \end{align*}$ So $\theta \in {\mathbb R}$ , meaning $\measuredangle YPO = \measuredangle YZP$ (in the directed sense). Thus, the circumcircle of $\triangle YPZ$ is tangent at $P$ to $\overline{OP}$ . Similarly, the circumcircle of $\triangle XPZ$ is tangent at $P$ to $\overline{OP}$ . This can only happen if $XYZP$ is concyclic, which concludes the problem.

Remark: The computation given here has at most eight terms at each step, so it is quite tractable.

If one does not realize the tangency point is $P$ , then he/she can still proceed in the following way. Search for a real number $t \in {\mathbb R}$ (equivalently $T \in \overline{OP}$ ) such that $\frac{y-t}{z-t} \div \frac{y-x}{z-x} \in {\mathbb R}$ . This will give a quadratic equation in $t$ . In order for line $\overline{OP}$ to be tangent, this quadratic equation must have a double root; in fact, one finds the quadratic $(b-c)(t^2-2t+1) = 0$ . I was able to carry out this longer approach by hand in about an hour.

Second solution by angle chasing (Eugene Lee) We prove several claims.

$[asy] size(12cm); pair O = origin; pair A = dir(110); pair B = dir(190); pair C = dir(350); pair P = dir(245); pair O_A = circumcenter(A, O, P); pair O_B = circumcenter(B, O, P); pair O_C = circumcenter(C, O, P); pair D = foot(O_A, B, C); pair E = foot(O_B, C, A); pair F = foot(O_C, A, B); pair X = extension(O_B, E, O_C, F); pair Y = extension(O_C, F, O_A, D); pair Z = extension(O_A, D, O_B, E); filldraw(unitcircle, invisible, deepcyan); draw(A--B--C--cycle, deepcyan); draw(O--P, deepcyan); filldraw(circumcircle(X, Y, Z), invisible, red); draw(circumcircle(A, O, P), lightblue+dashed); draw(circumcircle(B, O, P), lightblue+dashed); draw(circumcircle(C, O, P), lightblue+dashed); draw(X--Y--Z--cycle, red); draw(X--O_C, red); draw(A--P, deepgreen+dotted); draw(Y--P, deepgreen+dotted); draw(Z--C, deepgreen+dotted); draw(circumcircle(O_B, P, O_C), lightgreen+dashed); draw(circumcircle(O_C, P, O_A), lightgreen+dashed); draw(circumcircle(O_A, P, O_B), lightgreen+dashed); draw(O_A--O_C, orange); dot("$O$", O, dir(45)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$P$", P, dir(290)); dot("$O_A$", O_A, dir(O_A)); dot("$O_B$", O_B, dir(180)); dot("$O_C$", O_C, dir(345)); dot("$X$", X, dir(15)); dot("$Y$", Y, dir(90)); dot("$Z$", Z, dir(Z)); /* TSQ Source: !size(12cm); O = origin R45 A = dir 110 B = dir 190 C = dir 350 P = dir 245 R290 O_A = circumcenter A O P O_B = circumcenter B O P R180 O_C = circumcenter C O P R345 D := foot O_A B C E := foot O_B C A F := foot O_C A B X = extension O_B E O_C F R15 Y = extension O_C F O_A D R90 Z = extension O_A D O_B E unitcircle 0.1 lightcyan / deepcyan A--B--C--cycle deepcyan O--P deepcyan circumcircle X Y Z 0.1 yellow / red circumcircle A O P lightblue dashed circumcircle B O P lightblue dashed circumcircle C O P lightblue dashed X--Y--Z--cycle red X--O_C red A--P deepgreen dotted Y--P deepgreen dotted Z--C deepgreen dotted circumcircle O_B P O_C lightgreen dashed circumcircle O_C P O_A lightgreen dashed circumcircle O_A P O_B lightgreen dashed O_A--O_C orange */ [/asy]$

Claim: The points $X$ , $P$ , $O_B$ , $O_C$ are concyclic. Similarly, the points $Y$ , $P$ , $O_C$ , $O_A$ are concyclic, as are $Z$ , $P$ , $O_A$ , $O_B$ .

Proof. We have $\begin{align*} \measuredangle O_B X O_C &=\measuredangle CAB = \measuredangle CPB = \measuredangle CPO + \measuredangle OPB \\ &= \measuredangle OCP + \measuredangle PBO \\ &= (90^{\circ} - \measuredangle O_C P O) + (90^{\circ} - \measuredangle O P O_B) \\ &= -(\measuredangle O_C P O + \measuredangle O P O_B) = -\measuredangle O_C P O_B = \measuredangle O_B P O_C. \end{align*}$ $\blacksquare$

Similarly $Y, O_A, P, O_B$ , $Z, O_A, P, O_B$ concyclic.

Claim: Point $P$ lies on $(XYZ)$ .

Proof. Note that $P$ is the Miquel point of the complete quadrilateral $O_A O_B O_C X Y Z$ . $\blacksquare$

Claim: Lines $AX$ , $BY$ , $CZ$ concur at $P$ .

Proof. We show just $A$ , $X$ , $P$ collinear. Compute $\begin{align*} \measuredangle OPX &= \measuredangle OPO_B + \measuredangle O_BPX = (90^{\circ} - \measuredangle PBO) + \measuredangle O_B O_C X \\ &= (90^{\circ} - \measuredangle PBO) + \measuredangle(\overline{OP}, \overline{BA}) \\ &= \measuredangle BAP + \measuredangle(\overline{OP}, \overline{BA}) = \measuredangle OPA. \end{align*}$ $\blacksquare$

To finish, since the corresponding sides of $\triangle ABC$ and $\triangle XYZ$ are spirally similar, the last claim implies that a spiral similarity at the second intersection of $(XYZ)$ and $(ABC)$ maps $\triangle ABC$ and $\triangle XYZ$ through a $90^{\circ}$ rotation. This means the circles are orthogonal, as needed.

Remark: The second claim admits a proof by angle chase: since $O_A$ , $O_B$ , $O_C$ all lie on the perpendicular bisector of $OP$ , we get $\measuredangle YZP = \measuredangle O_A Z P = \measuredangle O_A O_B P = \measuredangle O_C O_B P = \measuredangle O_C X P = \measuredangle YXP.$
The last paragraph can also be replaced by angle chasing: $\begin{align*} \measuredangle PYX &= \measuredangle(\overline{PB}, \overline{XY}) = \measuredangle(\overline{PB}, \overline{AB})+90^{\circ} \\ &= \measuredangle PBA + 90^{\circ} = \measuredangle PCA + 90^{\circ} = 90^{\circ} - \measuredangle ACP \\ &= \measuredangle OPA = \measuredangle OPX \end{align*}$ as needed.

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sansae

119 posts

sansae

#7 h Jul 17, 2019, 3:27 PM • 2 Y

Y by centslordm, Adventure10

+) South Korean TST #3
I solved using complex numbers too, but much more messier than above posts

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MarkBcc168

1595 posts

MarkBcc168

#8 h Jul 17, 2019, 3:49 PM • 4 Y

Y by centslordm, Adventure10, Mango247, MS_asdfgzxcvb

Generalization: Let $O$ be the circumcenter of $\triangle ABC$ and $P$ be an arbitrary point. Let $O_A, O_B, O_C$ be the circumcenters of $\triangle AOP$ , $\triangle BOP$ , $\triangle COP$ respectively. Let $\ell_A, \ell_B, \ell_C$ be the lines through $O_A, O_B, O_C$ perpendicular to $BC, CA, AB$ respectively. Let $S$ be the circumcenter of triangle formed by $\ell_A, \ell_B, \ell_C$ . Prove that $\angle OPS = 90^{\circ}$ .

I have a prove but it's very long and not completely synthetic. I will spare the proof for a while.

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RC.

439 posts

RC.

#9 h Jul 18, 2019, 8:28 AM • 4 Y

Y by Aryan-23, centslordm, Adventure10, Mango247

Once you know that it is easier than G1, it is easier than G1.

$P$ is the required tangency point. Let $\Delta A_1B_1C_1$ be the required $\Delta$ with $A_1$ opposite to $l_a$ and so on. $\angle O_AOO_B = \angle AOB /2 = \angle C \Rightarrow O_APO_BC_1$ is cyclic similarly, $O_BO_CPA_1, O_APO_CB_1$ is cyclic. Combining the three, $A_1B_1C_1P$ is also cyclic. Let $PO \cap BC = X$ Now, $\angle OPC_1 = \angle OPO_B - \angle O_BPC_1 = \angle BCP - \angle PXC = \angle OPC = \angle PO_CO_B = \angle PA_1C_1 \Rightarrow \angle PA_1C_1 = \angle OPC_1 \Rightarrow \odot A_1B_1C_1P$ is tangent to $OP$ at $P$ .

This post has been edited 1 time. Last edited by RC., Jul 18, 2019, 8:30 AM

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MarkBcc168

1595 posts

MarkBcc168

#10 h Jul 27, 2019, 10:25 AM • 6 Y

Y by ABCCBA, Kanep, centslordm, Adventure10, Mango247, MS_asdfgzxcvb

Here is my solution to the generalization. Hopefully someone will find a simpler proof, especially to the main lemma.

MarkBcc168 wrote:

Lemma: Let $\triangle ABC$ be a triangle and $P$ be an arbitrary point. A line $\ell$ pass through $P$ intersect sides $BC, CA, AB$ at $D,E,F$ . Let the lines through $D,E,F$ parallel to $AP, BP, CP$ form the triangle $XYZ$ ( $D\in YZ$ , $E\in XZ$ ). Let $A', B', C'$ be reflections of $A,B,C$ across $\ell$ . Then lines $A'X, B'Y, C'Z$ and $\ell$ are concurrent.

Proof: Fix line $\ell$ and animate $P$ along $\ell$ . First, we claim that $X$ move along a fixed line. To prove that, let $A_1$ be the second intersection of $\odot(BPC)$ and $\ell$ . Then $\measuredangle XEF = \measuredangle BPA_1 = \measuredangle BCA_1$ or $\triangle XEF\stackrel{+}{\sim}\triangle A_1CB$ .

This means of a spiral similarity $\mathcal{T}$ which maps $C\to E$ , $B\to F$ takes line $\ell\to\ell_A$ , then $X\in\ell_A$ . As $\mathcal{T}$ is fixed, line $\ell_A$ is also fixed.

Similarly $Y,Z$ move along fixed lines $\ell_B, \ell_C$ . Now we are ready to use moving point method. Note that the following composition of mappings
$P\to{\infty}_{BP}\to X\to AX\cap \ell$ is projective. Similarly $P\to B'Y\cap\ell$ is projective. Thus it suffices to verify the lemma for three cases of $P$ .

Actually, by symmetry, it suffices to take $P=D$ . Then $Y,Z\in AD$ such that $YF\parallel ZE\parallel BC$ and $X$ is ${\infty}_{BC}$ . Let $P=B'Y\cap\ell$ . Then it suffices to show that $A'P\parallel BC$ . This can be rephrased to the following sub-lemma (applied on $\triangle ABD$ ).

Sub-lemma (The Special Case): Let $\triangle ABC$ be a triangle. Points $E, F$ are placed on line $AC, AB$ respectively such that $EF\parallel BC$ . Let $E'$ be the reflection of $E$ across $CF$ . Let the lines $BE'$ and $CF$ intersect at $P$ . Then $AP\parallel E'F$ .

Proof: Let $X$ be the point on $CF$ such that $E'P=E'X$ . Then $\triangle FE'X\sim\triangle CBP$ . Thus
$\frac{PE'}{PB} = \frac{XE'}{PB} = \frac{FE}{BC} = \frac{AF}{AB}$ or $AP\parallel E'F$ .

Back to the main problem.

Let $A_1=\ell_B\cap \ell_C$ . Define $B_1,C_1$ similarly.

Clearly $O_A, O_B, O_C$ lie on line $\ell_1$ perpendicular to $OP$ . Let line $OP\equiv\ell$ meet $BC, CA, AB$ at $D,E,F$ . Let $A', B', C'$ be the reflections of $A,B,C$ across $\ell$ . Then
$\measuredangle A'EF = -\measuredangle AEF = -\measuredangle(\ell_B, \ell_1) = -\measuredangle A_1O_BO_C$ or $\triangle A'EF\stackrel{-}{\sim} \triangle A_1O_BO_C$ . Now let $P'$ be the inverse of $P$ w.r.t. $\odot(ABC)$ . Let $X$ be the point such that $XE\parallel BP'$ and $XF\parallel CP'$ . Define $Y,Z$ similarly. Then
$\measuredangle XEF = \measuredangle BP'O = \measuredangle OBP = \measuredangle O_CO_BP$ or $\triangle XEF \stackrel{-}{\sim} \triangle PO_BO_C$ . Thus $\triangle A'EF\cup X\sim\triangle A_1O_BO_C\cup P$ . But by our result, $A'X, B'Y, C'Z$ meet at a point $Q\in\ell$ . Thus
$\measuredangle(AX, \ell) = \measuredangle(\ell, A'X) = -\measuredangle(\ell_1, A_1P) = \measuredangle(A_1P, \ell_1).$ or $A_1P\perp AQ$ . Thus $\triangle ABC\cup Q\cup O\sim\triangle A_1B_1C_1\cup P\cup S$ so we are done.

This post has been edited 1 time. Last edited by MarkBcc168, Jul 27, 2019, 10:25 AM

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buratinogigle

2356 posts

buratinogigle

#11 h Jul 27, 2019, 11:09 AM • 6 Y

Y by centslordm, geometrylover123, Adventure10, Mango247, Mango247, Mango247

More general problem. Let $ABC$ be a triangle inscribed in circle $(\omega)$ . $P$ and $Q$ are two any points. $DEF$ is circumcevian triangle of $Q$ with respect to $ABC$ . $J$ , $K$ , $L$ are circumcenters of triangles $APQ$ , $BPQ$ , $CPQ$ respectively. The lines $\ell_A, \ell_B, \ell_C$ pass through $J$ , $K$ , $L$ and are perpendicular to $EF$ , $FD$ , $DE$ respectively. Lines $\ell_A, \ell_B, \ell_C$ bound triangle $XYZ$ . Let $R$ be the point such that $\triangle XYZ\cup R\sim\triangle DEF\cup Q$ . Prove that $PR\perp PQ$ .

Attachments:

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MarkBcc168

1595 posts

MarkBcc168

#12 h Jul 27, 2019, 1:01 PM • 2 Y

Y by centslordm, Adventure10

If I'm not mistaken, my solution works for buratinogiggle's generalization. There is some significant modification on the last part (which you need to apply the lemma on $\triangle DEF$ instead). I will left the modification to readers.

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mastermind.hk16

143 posts

mastermind.hk16

#13 h Aug 7, 2019, 8:56 PM • 2 Y

Y by centslordm, Adventure10

Let $D,E,F$ be the feet of perpendiculars from $O_A,O_B,O_C$ to $BC,CA,AB$ respectively.
Let $O_AD \cap O_BE = Z \ O_BE \cap O_CF = X \ O_CF \cap O_AD = Y$ .

Now we will prove the following
Claim: $A,X,P$ are collinear.
Let $X_1 = AP \cap O_BE$ . Then $\angle O_BX_1P = \angle AX_1E =90^{\circ} - \angle X_1AE = 90^{\circ} - \angle PAC = \angle OCP = \angle O_BO_CP \longrightarrow (X_1O_BPO_C)$ is cyclic.
Let $X_2 = AP \cap O_CF$ . Analogously, $(X_2O_BPO_C)$ is cyclic. So $X_2 =X_1$ and our claim is proved.

Next, it is easy to see that $P \in (XYZ)$ . Indeed,
$\angle XPY = 180^{\circ}- \angle APB =180^{\circ} -\angle ACB = \angle ACD = \angle ECD = 180^{\circ} - \angle EZD = 180^{\circ} -\angle XZY$

Now our claim also implies $B,Y,P$ are collinear and $C,Z,P$ are collinear. Now,
$\angle CED = \angle CZD = \angle PZY = \angle PXY =\angle AXF =\angle AEF \longrightarrow D,E,F$ are collinear.
Lastly, $\angle OPX = \angle OPA = 90^{\circ} - \angle ABP =90^{\circ} -\angle ACZ =90^{\circ} -\angle ECZ = \angle EZC = \angle XZP$ So, $OP$ is tangent to $(XYZ)$ and we are done.

This post has been edited 5 times. Last edited by mastermind.hk16, Aug 14, 2019, 1:01 PM

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Wizard_32

1566 posts

Wizard_32

#14 h Dec 16, 2019, 9:24 AM • 4 Y

Y by amar_04, Aryan-23, centslordm, Adventure10

In fact a shorter solution using angel chasing is possible.

Neothehero wrote:

Let $\ell_B \cap \ell_C=X.$ Define $Y,Z$ similarly. Call the circle $XYZ$ as $\omega.$ Let $S$ be the center of $\omega.$ Also, $K=\ell_A \cap BC,$ and $L,M$ are similarly defined. We show that $\omega$ is tangent to $OP$ at $P.$

Claim: The point $P$ lies on $\omega.$
Proof: Clearly,
$\begin{align*} \measuredangle O_CPO_A=\measuredangle O_CPO+\measuredangle OPO_A &=(\pi/2-\measuredangle OCP)+(\pi/2-\measuredangle PAO) \\ &=\measuredangle PAC+\measuredangle ACP \\ &=\measuredangle ABC =\measuredangle KYL \end{align*}$ Hence, $PO_AYO_C$ is cyclic. Similarly, $O_AO_BZP$ is cyclic. Thus, we are done by Miquel on $XYO_AO_B.$ $\square$
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -13.0711337058945, xmax = 4.984202470977303, ymin = -4.693469917932022, ymax = 11.753212636798287; /* image dimensions */ pen ffqqtt = rgb(1,0,0.2); pen wwffqq = rgb(0.4,1,0); pen qqzzcc = rgb(0,0.6,0.8); pen qqccqq = rgb(0,0.8,0); draw((-6,1)--(-5,-2)--(1,-2)--cycle, linewidth(0.4) + ffqqtt); draw((-6.653718180703615,2.8812662866248995)--(-3.8397045330106034,2.7568789372060967)--(-3.839704533010603,0.43822317183382453)--(-8.344242012714654,-1.0632893214008594)--cycle, linewidth(0.4) + qqzzcc); draw((-3.330380703997218,-2)--(-3.330380703997218,-1.4906761709866152)--(-3.839704533010603,-1.4906761709866152)--(-3.839704533010603,-2)--cycle, linewidth(0.4) + qqccqq); 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draw((-3.8397045330106034,2.7568789372060967)--(2.3018417046366646,2.4854052510495808), linewidth(0.4)); draw((2.3018417046366646,2.4854052510495808)--(-3.839704533010603,0.43822317183382453), linewidth(0.4)); draw((-3.839704533010603,0.43822317183382453)--(-3.839704533010603,-2), linewidth(0.4)); draw((-6,1)--(-7.233427290924139,1.528611696110345), linewidth(0.4)); /* dots and labels */ dot((-6,1),linewidth(4pt) + dotstyle); label("$A$", (-6.372411906230626,1.3089689706556822), NE * labelscalefactor); dot((-5,-2),linewidth(4pt) + dotstyle); label("$B$", (-5.412021684056594,-2.4365528958230454), NE * labelscalefactor); dot((1,-2),linewidth(4pt) + dotstyle); label("$C$", (1.1666513378355259,-2.5085821624860976), NE * labelscalefactor); dot((-1.8227489189052903,4.676615938834815),linewidth(4pt) + dotstyle); label("$P$", (-1.7385290842409213,4.862412792699603), NE * labelscalefactor); dot((-2,0.6666666666666666),linewidth(4pt) + dotstyle); label("$O$", (-2.122685173110534,0.08447143738379048), NE * labelscalefactor); 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Thus, it suffices to show that $\measuredangle SPO=\pi/2.$ Now,

Now,
$\measuredangle ZPX=\measuredangle ZYX=\measuredangle KBL=\measuredangle CPA$ and hence $Z,P,C$ are collinear. Similarly, $\{P,A,X\}$ and $\{P,Y,B\}$ are collinear triples. Thus
$\measuredangle OPS=\measuredangle OPY+\measuredangle YPS=\measuredangle OPB+(\pi/2-\measuredangle PZY)=\measuredangle OPB+\measuredangle KCP=\pi/2$ Thus we are done. $\blacksquare$

This post has been edited 4 times. Last edited by Wizard_32, Dec 16, 2019, 11:42 AM

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yayups

1614 posts

yayups

#15 h Dec 22, 2019, 12:21 AM • 3 Y

Y by Wizard_32, centslordm, Adventure10

Let $XYZ$ be the triangle formed by $\ell_A$ , $\ell_B$ , $\ell_C$ . Here is the claim which allows the problem to fall to straight angle chasing.

Claim: The points $P,A,X$ are collinear (and symmetric variants)

Proof: We'll complex bash with $(ABC)$ as the unit circle. Note that $O_A$ is the intersection of the perpendicular bisector of $OA$ and $OP$ , so it is $1/2$ of the pole of $AP$ . This means that $O_A=\frac{ap}{a+p}.$ We'll now find $x$ . Note that $XO_B\perp AC$ , so the foot from $X$ to $AC$ is the foot from $O_B$ to $AC$ , so $\frac{1}{2}(x+a+c-ac\bar{x})=\frac{1}{2}(O_B+a+c-ac\bar{O}_B),$ so $ac(\bar{x}-\bar{O}_B)=x-O_B$ , and similarly $ab(\bar{x}-\bar{O}_C)=x-O_C$ . Setting the value for $x$ we get from these two equal, we have that $\begin{align*} a(b-c)\bar{x}&=a(b\bar{O}_C-c\bar{O}_B)+O_B-O_C \\ &= a\left(\frac{b}{c+p}-\frac{c}{b+p}\right)+\frac{bp}{b+p}-\frac{cp}{c+p} \\ &= \frac{a[b^2+bp-c^2-cp]+bp(c+p)-cp(b+p)}{(b+p)(c+p)} \\ &= \frac{(b-c)[ap+a(b+c)+p^2]}{(b+p)(c+p)}, \end{align*}$ so $\boxed{\bar{x}=\frac{ap+a(b+c)+p^2}{a(b+p)(c+p)}}.$ We also compute $\begin{align*} x &= \frac{\frac{1}{ap}+\frac{b+c}{bca}+\frac{1}{p^2}}{\frac{1}{a}\frac{b+p}{bp}\frac{c+p}{cp}} \\ &= \frac{bcp+p^2(b+c)+abc}{(b+p)(c+p)}. \end{align*}$ To verify $X,A,P$ collinear, all we have to show is that $x+ap\bar{x}=a+p$ . Indeed, $\begin{align*} x+ap\bar{x} &= \frac{bcp+p^2(b+c)+abc}{(b+p)(c+p)}+\frac{ap^2+ap(b+c)+p^3}{(b+p)(c+p)} \\ &= \frac{(a+p)(b+p)(c+p)}{(b+p)(c+p)} \\ &= a+p, \end{align*}$ completing the proof of the claim. $\blacksquare$

We're now in a position to solve the problem. We'll be using directed angles mod $\pi$ throughout. We see that $\triangle XYZ\stackrel{+}{\sim}\triangle ABC$ due to the perpendicularity conditions. Note that $\angle YPX = \angle BPA = \angle BCA = \angle YZX,$ so $P\in(XYZ)$ . To show that $OP$ is tangent to $(PXY)$ , it suffices to show that $\angle OPY=\angle PXY$ . Indeed, we have $\angle PXY = \angle AXY = \pi/2 - \angle BAP = \angle OPB = \angle OPY,$ as desired.

Remark: After drawing a nice diagram, I noticed that $P$ looks like it's on $(XYZ)$ , and later I noticed that $P,A,X$ (and variants) looked collinear. I tried for a bit doing synthetic stuff from here, but completely missed all the key observations (mainly that $O_BO_CXP$ cyclic). Instead, I saw that if we had $P,A,X$ collinear (and variants), then all the angles are computable and the problem has to fall to angle chase. This was a much more modest goal, and I saw that at most 10 minutes of complex bashing would show this. Hence, the above solution.

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Physicsknight

641 posts

Physicsknight

#16 h Feb 5, 2020, 7:09 AM • 2 Y

Y by centslordm, Adventure10

Let $E=\ell_A\cap\ell_C, D=\ell_B\cap\ell_A, F=\ell_B\cap\ell_C$ .
$\widehat {O_AEO_C}=\widehat {FED}=\widehat {ABC}$
$\widehat {O_APO_C}=\widehat {O_APO}+\widehat {O_CPO_B}=90^{\circ}-\widehat {OAP}+90^{\circ}-\widehat {OPC}=\widehat {ABC}\implies O_AEPO_C$ is cyclic. So, $\odot (O_BFPO_A)=\odot (O_CFBO_B)$ .
Let $\mathcal {P}$ be the Miquel point of $(DFO_AO_BFO_C)$ . $D,E,F,\mathcal {P}$ are cyclic.
$\widehat{FO_AO_B}=90^{\circ}-\widehat {AO_AO_B}=\widehat {AO_BP}-\widehat {O_APO}=\widehat {O_APO}+\widehat {OPC}=\widehat{O_APC}\implies\overline {F,P,C}=\overline{D,P,A}=\overline{E,P,B}$
$\widehat {OPD}=\widehat {OAP}=\widehat {PMO}=\widehat {PO_AO_C}=\widehat {PEO_C}=\widehat {DFD}$

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Aryan-23

558 posts

Aryan-23

#17 h May 1, 2020, 3:24 PM • 4 Y

Y by amar_04, GeoMetrix, AlastorMoody, centslordm

Thanks

We present a hybrid solution which is synthetic at heart, but uses complex numbers to prove a crucial claim.

Let $L_A,L_B,L_C$ be $\ell_B\cap\ell_C,\ell_A\cap\ell_C,\ell_A\cap\ell_B$ .

Claim 1:- $O_B,O_C,L_A,P$ are concyclic.

Proof:- Note that $\measuredangle{OBPO_C}=\measuredangle{O_BOO_C}=\pi-(\measuredangle{OBP}+\measuredangle{OCP})=A$

Claim 2:- $P$ lies on $(L_AL_BL_C)$

Proof:- Note that $O_A,O_B,O_C$ lie on the Perpendicular bisector of $OP$ . $\measuredangle{PL_CL_A}=\measuredangle{PL_CO_B}=\measuredangle{PO_AO_B}=\measuredangle{PO_AO_C}=\measuredangle{PL_BO_C}=\measuredangle{PL_BL_A}$

We will prove that $OP$ is tangent to $(L_AL_BL_CP)$ .

Claim 3:- $A,L_A,P$ are collinear.

The Proof is by Complex numbers with $(ABC)$ as the unit circle and $A=a,B=b,C=c$ and $P=1$ . So $OP$ is the real axis.

By Circumcenter formula $O_A=\frac{a}{a+1}$ (and cyclic variants),so we can compute $l_a$ now :-

Note that $O_CL_A\perp AB$ . So we have: $\frac{o_c-\ell_a}{a-b}+\overline{\frac{o_c-\ell_a}{a-b}}=0\implies \ell_a-ab\overline{\ell_a}=o_c-ab\overline{o_c}$ . Similarly $\ell_a-ac\overline{\ell_a}=o_b-ac\overline{o_b}$ . So from the two equations we get:- $(b-c)\ell_a=abc(\overline{o_c-o_b})+(bo_b-co_c)$ Note $\overline{o_c-o_b}=\frac{b-c}{(b+1)(c+1)}$ and $bo_b-co_c=\frac{b^2}{b+1}-\frac{c^2}{c+1}=\frac{(b+c+bc)(b-c)}{(b+1)(c+1)}$ . So $\ell_a=\frac{b+c+bc}{(b+1)(c+1)}+\frac{abc}{(b+1)(c+1)}\implies \ell_a=\frac{bc+b+c+abc}{(b+1)(c+1)}$ and $\overline{\ell_a}=\frac{a+ab+ac+1}{a(b+1)(c+1)}$ . So we need $A,L_A,P$ collinear $\implies \ell_a+a\overline{\ell_a}=\frac{(a+1)(b+1)(c+1)}{(b+1)(c+1)}=a+1$
Which is exactly what we wanted by chord formula

Finish:- Let $\measuredangle{OAP}=\theta=\measuredangle{OPA}=\measuredangle{OPL_A}$ . Now $\measuredangle{PL_CL_A}=\measuredangle{PO_AO_B}=\measuredangle{(PO_A,\ell)}=\frac{1}{2}\cdot 2(\measuredangle{PAO})=\theta$ . So, $\measuredangle{OPL_A}=\measuredangle{PL_CL_A}$ . Done!

This post has been edited 2 times. Last edited by Aryan-23, Jul 11, 2020, 11:40 AM

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mathsworm

765 posts

mathsworm

#18 h May 13, 2020, 7:15 AM • 1 Y

Y by centslordm

A moving points + angle chase solution (which is hopefully correct)
Let the triangle formed by the lines $l_i$ be $XYZ$ with $X$ being the vertex corresponding to $A$ .
Animate $P$ projectvely on $(ABC)$ . Then, $O_A$ , $O_B$ , $O_C$ move projectively on the perpendicular bisectors of $OA,OB,OC$ . Lines $l_i$ move projectively on the pencil of lines through the points at infinity along the lines perpendicular to $AB,BC,CA$ . Thus, $X,Y,Z$ move with degree $2$ each.
To show that $A,P,X$ are collinear, we need to show this for $5$ positions of $P$ . We can show this for $A,B,C$ and their antipodes.

Also, it is easy to see that $\triangle XYZ ~ \triangle ABC$ .
So, $\angle ZPX=\angle CPX=\angle CPA=\angle CBA= \angle ZYX$ (where the angles are directed) and hence, $P$ lies on $(XYZ)$ .
$\angle PYX=\angle PYZ+\angle ZYX =90^{\circ} + \angle PBC +\angle CBA=90^{\circ} + \angle PBA =90^{\circ} + \angle PCA =\angle OPX$ Hence done.

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nukelauncher

354 posts

nukelauncher

#19 h Jun 15, 2020, 10:38 PM • 2 Y

Y by mfang92, centslordm

Solution with mfang92, jclash

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ducktility

20 posts

ducktility

#20 h Jul 11, 2020, 11:34 AM • 1 Y

Y by centslordm

Neothehero wrote:

Solution : Let $A_1A_2A_3$ be the required triangle such that $A_1$ doesn't lie on $\ell _A$ and $A_2$ doesn't lie on $\ell _B$ and $A_3$ doesn't lie on $\ell _C$ . Now $m\angle O_AOO_B = m\angle C \implies O_APO_BA_3$ is cyclic and in a similar fashion, we have that $O_BO_CPA_1$ and $O_AA_2O_CP$ are also cyclic. We all together get $A_1A_2A_3P$ is cyclic. Now, if $X = PO \cap BC$ , then $m\angle OPA_3 = m\angle BCP - m\angle PXC$ $= m\angle OPC = m\angle PO_CO_B = m\angle PA_1A_3 \implies m\angle PA_1A_3 = m\angle OPA_3$ which gives us the desired result

This post has been edited 5 times. Last edited by ducktility, Jul 11, 2020, 11:38 AM

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richrow12

411 posts

richrow12

#21 h Jul 23, 2020, 3:37 PM • 1 Y

Y by centslordm

yayups wrote:

Let $XYZ$ be the triangle formed by $\ell_A$ , $\ell_B$ , $\ell_C$ . Here is the claim which allows the problem to fall to straight angle chasing.

We'll complex bash with $(ABC)$ as the unit circle. Note that $O_A$ is the intersection of the perpendicular bisector of $OA$ and $OP$ , so it is $1/2$ of the pole of $AP$ . This means that $O_A=\frac{ap}{a+p}.$ We'll now find $x$ . Note that $XO_B\perp AC$ , so the foot from $X$ to $AC$ is the foot from $O_B$ to $AC$ , so $\frac{1}{2}(x+a+c-ac\bar{x})=\frac{1}{2}(O_B+a+c-ac\bar{O}_B),$ so $ac(\bar{x}-\bar{O}_B)=x-O_B$ , and similarly $ab(\bar{x}-\bar{O}_C)=x-O_C$ . Setting the value for $x$ we get from these two equal, we have that $\begin{align*} a(b-c)\bar{x}&=a(b\bar{O}_C-c\bar{O}_B)+O_B-O_C \\ &= a\left(\frac{b}{c+p}-\frac{c}{b+p}\right)+\frac{bp}{b+p}-\frac{cp}{c+p} \\ &= \frac{a[b^2+bp-c^2-cp]+bp(c+p)-cp(b+p)}{(b+p)(c+p)} \\ &= \frac{(b-c)[ap+a(b+c)+p^2]}{(b+p)(c+p)}, \end{align*}$ so $\boxed{\bar{x}=\frac{ap+a(b+c)+p^2}{a(b+p)(c+p)}}.$ We also compute $\begin{align*} x &= \frac{\frac{1}{ap}+\frac{b+c}{bca}+\frac{1}{p^2}}{\frac{1}{a}\frac{b+p}{bp}\frac{c+p}{cp}} \\ &= \frac{bcp+p^2(b+c)+abc}{(b+p)(c+p)}. \end{align*}$

After obtaining expressions for $X, Y, Z$ one can notice that, for example
$x=\frac{bcp+p^2(b+c)+abc}{(b+p)(c+p)}=\frac{p(b+p)(c+p)+abc-p^3}{(b+p)(c+p)}=p+\frac{abc-p^3}{(b+p)(c+p)}= p+\frac{abc-p^3}{(a+p)(b+p)(c+p)}\cdot p+\frac{abc-p^3}{(a+p)(b+p)(c+p)}\cdot a.$ Let $s=p+\frac{abc-p^3}{(a+p)(b+p)(c+p)}\cdot p$ and $t=\frac{abc-p^3}{(a+p)(b+p)(c+p)}$ (note that this expressions are symmetric of $a,b,c$ ). Then, the triangle $XYZ$ is the image of initial triangle $ABC$ under the map $f\colon x\mapsto s+tz$ ( $f$ is a spiral similarity or translation). Hence, circumcenter of $XYZ$ is $f(0)=s$ and circumradius of $(XYZ)$ is equal to $|t|$ . Moreover, distance between points $p$ and $s$ is exactly $|pt|=|t|$ . So, in order to prove tangency, it's sufficient to prove that $OP$ is perpendicular to $pt$ , which is quite easy:
$\frac{p}{\overline{p}}=p^2~\text{and}~\frac{pt}{\overline{pt}}=\frac{-abcp^3}{ap\cdot bp\cdot cp}\cdot p^2=-p^2,$ as desired.

Remark. This solution uses very little geometry and some simple algebraic transforamations.

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pad

1671 posts

pad

#22 h Nov 3, 2020, 4:48 AM • 1 Y

Y by centslordm

Diagram

Let $X=\ell_B\cap \ell_C$ , and similarly define $Y$ and $Z$ . We will show that $(XYZ)$ is tangent to $OP$ at $P$ .

First, we claim $\triangle XYZ\sim \triangle ABC$ . Indeed, $\angle XYZ= \angle (XY,YZ) = \angle (\perp AB, \perp BC) = \angle (AB,BC) = \angle ABC,$ and cyclic variants, proving the similarity.

Claim: $XO_BPP_C$ is cyclic (and cyclic variants).

Proof: First note $\begin{align*} \angle O_APO &= 90^\circ-\tfrac12 \angle OO_AP = 90^\circ-\angle OAP = 90-(\tfrac{180-\angle AOP}{2}) = \tfrac12\angle AOP, \end{align*}$ and similar cyclic variants. We have $\begin{align*} \angle O_BXO_C &= \angle ZXY = \angle BAC, \\ \angle O_BPO_C &= \angle O_BPO + \angle O_CPO =\tfrac12 \angle BOP+\tfrac12 COP \\ &= \tfrac12 (360^\circ-\angle BOC) =\angle BAC, \end{align*}$ proving the claim. $\blacksquare$

Claim: $P,A,X$ are collinear (and cyclic variants).

Proof: Note that $O_AO_BO_C$ collinear since they lie on the perpendicular bisector of $OP$ . Hence $\begin{align*} \angle PXO_C &= \angle PO_BO_C= \tfrac12 \angle PO_BO=\angle PBO \\ &= 90-\tfrac12 \angle BOP = 90^\circ -\angle PCB,\\ \angle AXO_C &= \angle AXY = 90^\circ - \angle XAB = \angle PAB - 90^\circ, \end{align*}$ which are clearly equal. $\blacksquare$

Now, $P\in (XYZ)$ since $\begin{align*} \angle PZY &= 90^\circ-\angle PCB \qquad \text{(since $\overline{ZPC}$ collinear)} \\ \angle PXY &= 90^\circ-\angle XAB = \angle PAB-90^\circ, \end{align*}$ which are equal since $P\in (ABC)$ . We now show $(XYZ)$ tangent to $OP$ at $P$ . We have $\begin{align*} \angle OPX &= \angle OPA = 90^\circ-\tfrac12 \widehat{AP} = 90^\circ-\angle ACP, \\ \angle XZP &= \angle XZY+\angle YZP = \angle ACB+(90^\circ-\angle PCB) = 90^\circ-\angle ACP, \end{align*}$ proving the tangency.

Remarks

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WYXkk

6 posts

WYXkk

#23 h Nov 14, 2020, 8:54 AM • 3 Y

Y by centslordm, Mango247, Mango247

Maybe too easy for a G7.
Solution

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mathaddiction

308 posts

mathaddiction

#24 h Dec 25, 2020, 8:26 AM • 2 Y

Y by centslordm, Mango247

$[asy] size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.724555607032263, xmax = 1.5754642722882315, ymin = -1.6167851463147418, ymax = 1.233769659836653; /* image dimensions */pen qqwuqq = rgb(0,0.39215686274509803,0); pen zzttff = rgb(0.6,0.2,1); pen fuqqzz = rgb(0.9568627450980393,0,0.6); pen ffvvqq = rgb(1,0.3333333333333333,0); /* draw figures */draw(circle((0,0), 1), linewidth(0.8) + blue); draw((-0.47283605161038517,-0.881150423195439)--(-0.4450539838751519,0.8955037417213041), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-0.47283605161038517,-0.881150423195439)--(-1.105389951335395,-0.253180878805536), linewidth(0.8) + linetype("4 4") + qqwuqq); draw((-1.1525563272989778,-1.0398503296756219)--(0.8141392612587833,-0.5806696679498616), linewidth(0.8) + linetype("4 4") + qqwuqq); draw(circle((-0.882555286012392,-0.6612900906532113), 0.4649821683296528), linewidth(0.8) + blue); draw(circle((-0.9915389138274326,-0.3662067951476439), 0.16042729352143456), linewidth(0.8) + blue); draw(circle((-0.4556079991573275,0.22057790875705574), 0.6750083460528921), linewidth(0.8) + blue); draw((0,0)--(-0.47283605161038517,-0.881150423195439), linewidth(0.8)); draw((-1.1180791812941542,-0.4648195591088224)--(0.21254118456735152,0.12459387086998551), linewidth(0.8)); draw((-1.1525563272989778,-1.0398503296756219)--(-1.0891887874249246,0.017031967878854862), linewidth(0.8) + zzttff); draw((-0.4450539838751519,0.8955037417213041)--(-0.8776878763194698,-0.47923271148975133), linewidth(0.8) + fuqqzz); draw((-1.1180791812941542,-0.4648195591088224)--(0.8141392612587833,-0.5806696679498616), linewidth(0.8) + fuqqzz); draw((0.8141392612587833,-0.5806696679498616)--(-0.4450539838751519,0.8955037417213041), linewidth(0.8) + fuqqzz); draw((-1.0891887874249246,0.017031967878854862)--(0.15146104647116193,-0.6487158470778117), linewidth(0.8)); draw((-1.1525563272989778,-1.0398503296756219)--(0.21254118456735152,0.12459387086998551), linewidth(0.8) + zzttff); draw((-1.105389951335395,-0.253180878805536)--(0.15146104647116193,-0.6487158470778117), linewidth(0.8) + zzttff); draw((-0.9784863924287199,-0.20631136622064544)--(0.4450539838751519,-0.8955037417213041), linewidth(0.8) + linetype("4 4") + ffvvqq); draw((0,0)--(-0.9784863924287199,-0.20631136622064544), linewidth(0.8)); /* dots and labels */dot((0,0),dotstyle); label("$O$", (-0.06567464619104664,0.03913880746615159), NE * labelscalefactor); dot((-0.4450539838751519,0.8955037417213041),dotstyle); label("$A$", (-0.5092257052395002,0.970596031467904), NE * labelscalefactor); dot((-0.8776878763194698,-0.47923271148975133),dotstyle); label("$B$", (-0.9025076442624623,-0.5700046469603912), NE * labelscalefactor); dot((0.8141392612587833,-0.5806696679498616),dotstyle); label("$C$", (0.8983096554742591,-0.6350588022874977), NE * labelscalefactor); dot((-0.47283605161038517,-0.881150423195439),dotstyle); label("$P$", (-0.4678276063949779,-0.9869426424659374), NE * labelscalefactor); dot((-1.0891887874249246,0.017031967878854862),linewidth(4pt) + dotstyle); label("$O_{A}$", (-1.0769710608215208,0.04209581452647461), NE * labelscalefactor); dot((-0.3675337929660405,-0.3702168936349633),linewidth(4pt) + dotstyle); label("$O_{B}$", (-0.41164447224884043,-0.30091700447099606), NE * labelscalefactor); dot((0.15146104647116193,-0.6487158470778117),linewidth(4pt) + dotstyle); label("$O_{C}$", (0.1620148974538262,-0.6261877811065286), NE * labelscalefactor); dot((-0.4661620144395024,-0.4543479242071927),linewidth(4pt) + dotstyle); label("$X_{A}$", (-0.5772368676269297,-0.56946640690940844), NE * labelscalefactor); dot((-1.105389951335395,-0.253180878805536),linewidth(4pt) + dotstyle); label("$X_{B}$", (-1.1982083502947647,-0.1855937291183982), NE * labelscalefactor); dot((-1.1525563272989778,-1.0398503296756219),linewidth(4pt) + dotstyle); label("$X_{C}$", (-1.2603054985615483,-1.037211762491429), NE * labelscalefactor); dot((-1.1180791812941542,-0.4648195591088224),linewidth(4pt) + dotstyle); label("$F_A$", (-1.106541131424751,-0.4398963363061782), NE * labelscalefactor); dot((-0.8334782621247745,-0.33875234396129433),linewidth(4pt) + dotstyle); label("$F_C$", (-0.7871843689098644,-0.2979599974106731), NE * labelscalefactor); dot((0.21254118456735152,0.12459387086998551),linewidth(4pt) + dotstyle); label("$F_B$", (0.27142415868577807,0.11010697691390414), NE * labelscalefactor); dot((-0.9784863924287199,-0.20631136622064544),linewidth(4pt) + dotstyle); label("$Q$", (-0.976432820770538,-0.16489467969613703), NE * labelscalefactor); dot((0.4450539838751519,-0.8955037417213041),dotstyle); label("$A'$", (0.45771560348612855,-0.8657053529926936), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$
Suppose $l_B\cap l_C=X_A$ , $l_C\cap l_A=X_B$ , $l_A\cap l_B=X_C$ . Let $F_A,F_B,F_C$ be the projection of $O_A,O_B,O_C$ on $BC,CA,AB$ respectively.
CLAIM. $\triangle AF_BF_C\sim \triangle OO_BO_C$
Proof.
We have $\angle AF_BF_C=\frac{1}{2}\angle BOC=\angle F_BAF_C$
Now we use complex number with $(ABC)$ as the unit circle, we have
$O_B=\frac{\begin{vmatrix}b&1&1\\0&0&1\\ p&1&1\end{vmatrix}}{\begin{vmatrix}b&\frac{1}{b}&1\\0&0&1\\ p&\frac{1}{p}&1\end{vmatrix}}=\frac{bp}{b+p}$ similarly $O_C=\frac{cp}{c+p}$ , therefore,
$F_B=\frac{1}{2}\left(a+c+\frac{bp}{b+p}-\frac{ac}{b+p}\right)$ $F_C=\frac{1}{2}\left(a+b+\frac{cp}{c+p}-\frac{ab}{c+p}\right)$ Therefore,
$\begin{align*} \frac{AF_B}{AF_C}&=\frac{|b+\frac{cp}{c+p}-\frac{ab}{c+p}-a|}{|c+\frac{bp}{b+p}-\frac{ac}{b+p}-a|}\\ &=\frac{|c+p|}{|b+p|}\cdot\frac{|(b-a)(c+p)+cp-ab|}{|(c-a)(b+p)+bp-ac|}\\ &=\frac{|c+p|}{|b+p|}\\ &=\frac{OO_B}{OO_C}\\ \end{align*}$ as desired. $\blacksquare$

Corollary 1. $F_A,F_C,F_B$ are collinear.
Proof.
By Menelaus theorem,
$\frac{CF_A}{F_AB}\cdot\frac{BF_C}{F_CA}\cdot\frac{AF_B}{F_BC}=\frac{OO_A}{OO_B}\cdot\frac{OO_C}{OO_A}\cdot\frac{OO_B}{OO_C}=1$ $\blacksquare$

Corollary 2. $A,X_A,P$ are collinear.
Proof.
$\angle BAX_A=90^{\circ}-\angle AX_AF_C=90^{\circ}-\angle AF_BF_C=90^{\circ}-\angle OO_BO_C=\angle O_BOP=\angle BAP$ $\blacksquare$
Therefore, $\angle X_BX_AP=\angle F_AF_BC=180^{\circ}-\angle X_BX_CP$ Hence $P$ lies on $(X_AX_BX_C)$
Therefore, $\angle OPX_A=90^{\circ}-\angle ACP=\angle X_AX_CP$ which implies $OP$ is tangent to $(X_AX_VX_C)$ as desired.

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L567

1184 posts

L567

#25 h May 24, 2021, 7:01 AM • 1 Y

Y by centslordm

Very easy for a G7 if you use GeoGebra

. But fun problem to do!

Let $\ell_A \cap \ell_B = Z, \ell_B \cap \ell_C = Y, \ell_A \cap \ell_C = X$ . Let $R,S,T$ be the feet of perpendiculars from $O_A, O_B, O_C$ onto $BC, CA, AB$

Claim 1: $PYO_BO_C$ is cyclic

Proof: See that $ASYT$ is cyclic. So, $\angle O_BYO_C = \angle BAC$ and also, $\angle O_BPO_C = 180 - \angle OBP - \angle OCP = 180 - \angle OBC - \angle OCB - \angle PCB - \angle PBC = 2 \angle BAC - \angle BAC = \angle BAC$ and so we get $PYO_BO_C$ is cyclic. $\blacksquare$

Similarly, we get that $PZO_AO_B$ and $PXO_AO_C$ is cyclic.

Now, observe that in quadrilateral $O_AO_BYX$ , $P = (O_AO_BZ) \cap (O_BO_CY)$ which means $P$ is its miquel point and so $PXYZ$ is cyclic as well.

Claim 2: $Z,P,C$ are collinear

Proof: $\angle BPC = 180 - \angle BAC = 180 - \angle TYZ = 180 - \angle XYZ = 180 - \angle XPZ = 180 - \angle BPZ$ and so $Z,P,C$ are collinear. $\blacksquare$

Similarly, $X,B,P$ and $A,Y,P$ are collinear.

Now, to finish, see that $\angle YPO = \angle APO = \angle PAO = \frac{\angle OO_AP}{2} = \angle O_BO_AP = \angle O_BZP = \angle YZP$ and so $OP$ is tangent to $(XYZ)$ at $P$ and so we're done

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ppanther

160 posts

ppanther

#26 h Jul 15, 2021, 8:02 PM

Y by

Extension. Denote by $XYZ$ this triangle determined by the lines. Let $k$ be the line such that $XYZ$ is the paralogic triangle of $ABC$ with respect to $k$ (i.e. it passes through $YZ \cap BC$ , etc.). Prove that the nine-point center of $ABC$ lies on $k$ .

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Eyed

1065 posts

Eyed

#27 h Jul 19, 2021, 10:50 PM • 2 Y

Y by VulcanForge, tree_3

Solved with Kevin Wu, Nacho Cho, Isaac Zhu

We proceed with complex numbers. Let $(ABCP)$ be the unit circle, and $A = a, B = b, C = c, P = p$ . First, observe that
$o_{a} = \frac{ap}{a+p}, o_{b} = \frac{bp}{b+p}, o_{c} = \frac{cp}{c+p}$ Let $XYZ$ be the triangle determined by $\ell_{A}, \ell_{B}, \ell_{C}$ . I claim that $X$ lies on $PC$ . Observe that:
$\frac{x-o_{a}}{b-c} = -\overline{\left(\frac{x-o_{a}}{b-c}\right)} = \frac{bc(\overline{x} - \overline{o-{a}}}{b-c}\Rightarrow x-o_{a} = bc(\overline{x}-\overline{o_{a}})$ We similarly have $x - o_{b} = ac(\overline{x}-\overline{o_{b}} )$ . Now, subtracting these two, we get
$o_{a} - o_{b} = bc\overline{o_{a}} - ac\overline{o_{b}} + \overline{x}(ac - bc)$ $\frac{ap(b+p) - bp(a+p)}{(a+p)(b+p)} = \frac{bc(b+p) - ac(a+p)}{(a+p)(b+p)} + \overline{x}(ac - bc)$ $\overline{x}(a-b)(c)(a+p)(b+p) = (a-b)(p^{2} + ac + bc + cp) \Rightarrow \overline{x} = \frac{p^{2} + c(a+b+p)}{c(a+p)(b+p)}$ Now, observe that
$x = \overline{\left(\frac{p^{2} + c(a+b+p)}{c(a+p)(b+p)}\right)} = \frac{\frac{1}{p^{2}} + \frac{1}{c}(\frac{1}{a} + \frac{1}{b} + \frac{1}{p})}{\frac{1}{c}(\frac{a+p}{ap})(\frac{b+p}{bp})} \cdot \frac{p^{2}cab}{p^{2}cab} = \frac{abc +p(ab+ap+bp)}{(a+p)(b+p)}$ Now, we will show $C,P,X$ collinear. To prove this, we use the fact that $X$ lies on the chord if $x + pc\overline{x} = p+c$ . Plugging it in, we get
$\frac{abc + p(ab + ap + bp) + p(p^{2} + c(a+b+p))}{(a+p)(b+p)}$ $= \frac{abc + p^{3} + acp + bcp + p^{2}c + abp + ap^{2} + bp^{2}}{(a+p)(b+p)} = \frac{(a+p)(b+p)(c+p)}{(a+p)(b+p)} = p+c$ Therefore, $X, P,C$ are collinear. Similarly, $P,B,Y$ and $P,A,Z$ are collinear. Now, we get
$\angle ZYX = \angle ABC = 180 - \angle APC = \angle ZPX$ This means $(ZYPX)$ is cyclic. Finally, observe that
$\angle YPO = 90 - \angle PCB = \angle YXP$ Therefore, $(ZYXP)$ is tangent to $OP$ at $P$ . We conclude.

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khanhnx

1618 posts

khanhnx

#28 h Jul 20, 2021, 12:57 AM

Y by

Suppose $l_A, l_B, l_C$ bound the triangle $DEF$ . We have $(PO_B, PO_C) \equiv (OO_C, OO_B) \equiv (PC, PB) \equiv (DO_C, DO_B) \pmod \pi$ Then $P$ $\in$ $(DO_BO_C)$ . Similarly, we have $P$ is Miquel point of completed quadrilateral $DEO_AO_B.O_CF,$ which means $P$ $\in$ $(DEF)$ . We also have $(PD, PA) \equiv (PD, PO_B) + (PO_B, PA) \equiv (O_CD, O_CO_B) + (PO_B, PB) + (PB, PA) \equiv (AB, OP)$ $+ \dfrac{\pi}{2} + (OP, OB) + (PB, PA) \equiv (AB, OP) + \dfrac{\pi}{2} + (BP, BA) + (AP, AB) \equiv 0 \pmod \pi$ From this, we have $A, X, P$ are collinear. Therefore, we have $(PO, PD) \equiv (AP, AO) \equiv (O_AP, O_AO_B) \equiv (FP, FD) \pmod \pi$ or $OP$ touches at $(DEF)$ at $P$

This post has been edited 2 times. Last edited by khanhnx, Jul 20, 2021, 12:59 AM

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Mogmog8

1080 posts

Mogmog8

#29 h Jan 16, 2022, 12:31 AM • 2 Y

Y by centslordm, megarnie

Let $\ell_A,\ell_B,$ and $\ell_C$ form triangle $DEF.$

Claim: $A,D,$ and $P$ are collinear.
Proof. WLOG let $(ABC)$ be the unit circle. Notice $O_A=\frac{ap(\overline{a}-\overline{p})}{\overline{a}p-a\overline{p}}=\frac{ap}{a+p}.$ Also, $\overline{O_CD}\perp\overline{AB}$ and $\overline{O_BD}\perp\overline{AC}$ so $d-ab\overline{d}=o_c-ab\overline{o_c}$ and $d-ac\overline{d}=bo_b-ac\overline{o_b}.$ Multiplying the equations by $c$ and $b,$ respectively and subtracting yields $\begin{align*}(b-c)d&=bo_b-co_c-abc(\overline{o_b}-\overline{o_c})\\&=\frac{b^2p}{b+p}-\frac{c^2p}{c+p}-abc\left(\frac{1}{b+p}-\frac{1}{c+p}\right)\\&=\frac{b^2p(c+p)-c^2p(b+p)-abc(c-b)}{(b+p)(c+p)}\\&=\frac{p(b-c)(pb+pc+bc)+abc(b-c)}{(b+p)(c+p)}\end{align*}$ so $d=\frac{p(pb+pc+bc)+abc}{(b+p)(c+p)}.$ Hence, $\begin{align*}d+ap\overline{d}&=\frac{p(pb+pc+bc)+abc}{(b+p)(c+p)}+ap\cdot\frac{\frac{ac+ab+ap+p^2}{abcp^2}}{\frac{(b+p)(c+p)}{bcp^2}}\\&=\frac{p(pb+pc+bc)+abc+p(ac+ab+ap+p^2)}{(b+p)(c+p)}\\&=\frac{(a+p)(b+p)(c+p)}{(b+p)(c+p)}\\&=a+p\end{align*}$ and $D$ lies on $\overline{AP}.$ $\blacksquare$

Therefore, $\measuredangle DPE=\measuredangle APB=\measuredangle ACB=90-\measuredangle(\overline{FO_A},\overline{AC})=\measuredangle(\overline{FO_B},\overline{FO_A})=\measuredangle DFE$ so $P$ lies on $(DEF).$ Additionally, $\measuredangle FPO=\measuredangle CPO=90-\tfrac{1}{2}\widehat{CP}=90-\measuredangle PAC=\measuredangle FDA.$ $\square$

This post has been edited 1 time. Last edited by Mogmog8, Jan 16, 2022, 4:13 AM

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BVKRB-

322 posts

BVKRB-

#32 h Jan 30, 2022, 1:16 PM • 1 Y

Y by nathantareep

I solved this without Geogebra, and I consider this to be my greatest achievement till now

This solution doesn't use any miquel point, complex number stuff, it is just angel chase (simpler than in #2 and I would argue one of the most simplest in this thread)
A recent G7 for my $(169=100+69=13^2)^{th}$ Post? Let's do this!

This diagram is almost the same as the one I used when I drew it on paper!
$[asy]/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.564632340869464, xmax = 15.051238838636193, ymin = -4.935646511438435, ymax = 8.175723762645454; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); pen ccqqqq = rgb(0.8,0,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((4.4484063808077,-2.5391535307129325)--(4.4431514829389664,-2.176565577770285)--(4.080563529996319,-2.181820475639019)--(4.085818427865053,-2.5444084285816664)--cycle, linewidth(1) + qqwuqq); /* draw figures */ draw((1.28,4.27)--(-0.44,-2.61), linewidth(1) + qqwuqq); draw((-0.44,-2.61)--(7.84,-2.49), linewidth(1) + qqwuqq); draw((7.84,-2.49)--(1.28,4.27), linewidth(1) + qqwuqq); draw(circle((3.6627636363636364,0.01930909090909069), 4.872980171456166), linewidth(1) + ccqqqq); draw((3.9781883066508104,4.882069935201065)--(3.6627636363636364,0.01930909090909069), linewidth(1)); draw((3.6627636363636364,0.01930909090909069)--(-0.44,-2.61), linewidth(1) + qqwuqq); draw((-0.44,-2.61)--(3.9781883066508104,4.882069935201065), linewidth(1) + linetype("2 2") + ffvvqq); draw((1.28,4.27)--(3.6627636363636364,0.01930909090909069), linewidth(1) + qqwuqq); draw((3.6627636363636364,0.01930909090909069)--(7.84,-2.49), linewidth(1) + qqwuqq); draw((3.04011185687238,6.672826642536562)--(7.84,-2.49), linewidth(1) + qqwuqq); draw((0.34560291213323907,4.058036897345917)--(3.9781883066508104,4.882069935201065), linewidth(1) + linetype("2 2") + ffvvqq); draw((-0.9892525278321339,2.762674221876561)--(3.04011185687238,6.672826642536562), linewidth(1) + qqwuqq); draw((7.810161900580135,2.191897150234193)--(0.34560291213323907,4.058036897345917), linewidth(1) + linetype("2 2") + ffvvqq); draw((3.1006255348580316,2.497382861526589)--(3.04011185687238,6.672826642536562), linewidth(1)); draw(circle((2.039911296137204,5.007796946366772), 1.9423503934208743), linewidth(1) + linetype("2 2") + qqwuqq); draw((3.9781883066508104,4.882069935201065)--(4.085818427865053,-2.5444084285816664), linewidth(1)); draw(circle((1.1807267907781123,4.557718637465095), 2.8162021761819207), linewidth(1) + linetype("2 2") + blue); draw((-0.9892525278321339,2.762674221876561)--(7.810161900580135,2.191897150234193), linewidth(1) + qqwuqq); /* dots and labels */ dot((1.28,4.27),dotstyle); label("$A$", (1.3415661400504038,4.449154375748078), NE * labelscalefactor); dot((-0.44,-2.61),dotstyle); label("$B$", (-0.7952282331705735,-2.8501352031747653), NE * labelscalefactor); dot((7.84,-2.49),dotstyle); label("$C$", (8.02545893948562,-2.713380363288623), NE * labelscalefactor); dot((3.9781883066508104,4.882069935201065),dotstyle); label("$P$", (4.042474227801719,5.04745680024995), NE * labelscalefactor); dot((3.6627636363636364,0.01930909090909069),linewidth(4pt) + dotstyle); label("$O$", (3.734775838057898,0.15847127432036456), NE * labelscalefactor); dot((-0.9892525278321339,2.762674221876561),linewidth(4pt) + dotstyle); label("$O_B$", (-1.4277193676439828,2.43202048742748), NE * labelscalefactor); dot((3.1006255348580316,2.497382861526589),linewidth(4pt) + dotstyle); label("$O_A$", (2.9655298636983463,2.0388503227548207), NE * labelscalefactor); dot((7.810161900580135,2.191897150234193),linewidth(4pt) + dotstyle); label("$O_C$", (7.683571839770264,1.8166237079398395), NE * labelscalefactor); dot((0.34560291213323907,4.058036897345917),linewidth(4pt) + dotstyle); label("$Z$", (0.008206451160513944,4.004701146118116), NE * labelscalefactor); dot((3.04011185687238,6.672826642536562),linewidth(4pt) + dotstyle); label("$X$", (3.102284703584489,6.808175363784032), NE * labelscalefactor); dot((3.087943376601477,3.372451781228858),linewidth(4pt) + dotstyle); label("$Y$", (2.7945863138406684,3.577342271473921), NE * labelscalefactor); dot((4.085818427865053,-2.5444084285816664),linewidth(4pt) + dotstyle); label("$P_X$", (4.025379872815951,-2.918512623117836), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Notations for the intersection points are clear from my diagram and $P_X$ is the foot of $P$ on $BC$
Claim: $\triangle XYZ \sim \triangle CBA$
Proof: Notice that the perpendiculars from the circumcenters induce some cyclic quads one of which is { $\ell_a \cap BC,C,X,l_b \cap AC$ } which gives that $\measuredangle YXZ = \measuredangle BCA$ and similarly for the other vertices which establishes the claim $\blacksquare$
Claim: $XPO_AO_B$ is cyclic
Proof: $\angle O_BPO_A = \angle O_BOO_A = \angle O_BOP - \angle O_AOP = \frac{\angle BOP - \angle AOP}{2}= \frac{\angle BOA}{2}= \angle BCA=\angle ZXY=\angle O_BXO_B \ \blacksquare$ Similarly the other cyclic quads corresponding to the other circumcenters can be found.
Claim: $P \in \odot(XYZ)$
ProofTrivially $O_A-O_B-O_C$ because all of them lie on the perpendicular bisector. Using this and the above claim we have $\angle PYX = \angle PO_CO_A = \angle PO_CO_B=\angle PZX \implies P \in \odot(XYZ) \ \blacksquare$ Claim: $X-P-C$
Proof: $\angle PXO_A=\angle PXY=\angle PZY=\angle PZO_C=\angle PO_BO_C=\frac{\angle PO_BO}{2}=\frac{180^{\circ}-2\cdot \angle POO_B}{2}=90^{\circ}-\angle POO_B=90^{\circ}-\angle PCB=\angle CPP_X$ Now notice that $XO_A \parallel PP_X$ and $\angle PXO_A=\angle CPP_X$ which means $X-P-C$ and similarly $P-Y-B$ proving our claim $\blacksquare$
For the finish note that $PP_X$ is an altitude and $O$ is the circumcenter of $\triangle PBC$ which means those two lines are isogonal, and from a previous angle chase we know that $\angle PXY=\angle CPP_X=\angle BPO=\angle YPO \implies OP \ \text{is tangent to} \ \odot(XYZ) \ \blacksquare$

This post has been edited 1 time. Last edited by BVKRB-, Jan 30, 2022, 1:16 PM

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SerdarBozdag

892 posts

SerdarBozdag

#33 h Feb 7, 2022, 4:36 PM • 1 Y

Y by MS_asdfgzxcvb

Let $W'$ be the antipode of $W$ for any point on $(ABC)$ . Let $XYZ$ be the triangle formed by the three lines. From angle chasing $XYZ \sim ABC$ . Also $PO_CO \sim POC'$ and $PO_BO \sim POB' \implies PB'C' \sim PO_CO_B \implies P$ is the miquel point of the $\{l_a,l_b,l_c,O_A-O_B-O_C\} \implies PXYZ$ is cyclic. Also note that $P'BC \sim PB'C' \sim PO_CO_B \sim PYZ$ . Thus we have $ABC \cap P' \sim XYZ \cap P$ with $90^{\circ}$ rotation counter-clockwise. If $O_P$ is the center of $(XYZ)$ then $OP' \perp PO_P$ as desired.

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rama1728

800 posts

rama1728

#35 h Feb 10, 2022, 5:24 PM • 1 Y

Y by Periwinkle27

Neothehero wrote:

Absolutely beautiful geo! Solved with MathsCrazy

We begin by proving some claims.
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(20cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.833803243714863, xmax = 9.227857215833756, ymin = -2.845547934811449, ymax = 5.9708899723658035; /* image dimensions */ /* draw figures */ draw((0.21836396866421326,3.8883221775966788)--(-3.36,-0.96), linewidth(2)); draw((-3.36,-0.96)--(3.38,-1.02), linewidth(2)); draw((3.38,-1.02)--(0.21836396866421326,3.8883221775966788), linewidth(2)); draw(circle((0.02138589341213731,0.2890153599634323), 3.604692764937281), linewidth(2)); draw(circle((0.3400502291073863,2.076619307972976), 1.815784908457883), linewidth(2)); draw(circle((2.405158160631204,1.4419935792728382), 2.6479669552629654), linewidth(2)); draw(circle((-2.931677849389184,3.0820499608844165), 4.064680522622853), linewidth(2)); draw((xmin, 112.33333333333324*xmin-36.12235642842339)--(xmax, 112.33333333333324*xmax-36.12235642842339), linewidth(2)); /* line */ draw((xmin, 0.6441378371139966*xmin + 4.970454589904979)--(xmax, 0.6441378371139966*xmax + 4.970454589904979), linewidth(2)); /* line */ draw((xmin, -0.7380623311708246*xmin + 3.2171502181428373)--(xmax, -0.7380623311708246*xmax + 3.2171502181428373), linewidth(2)); /* line */ /* dots and labels */ dot((0.21836396866421326,3.8883221775966788),dotstyle); label("$A$", (0.2719187088593376,4.031831635185997), NE * labelscalefactor); dot((-3.36,-0.96),dotstyle); label("$B$", (-3.2992966459466606,-0.8227892377533768), NE * labelscalefactor); dot((3.38,-1.02),dotstyle); label("$C$", (3.438582324253719,-0.8785894776722202), NE * labelscalefactor); dot((0.02138589341213731,0.2890153599634323),linewidth(4pt) + dotstyle); label("$O$", (0.07661786914338459,0.40481604046117736), NE * labelscalefactor); dot((1.0802676387914532,3.7346763094611677),dotstyle); label("$P$", (1.1368224276014154,3.8783809754091774), NE * labelscalefactor); dot((0.3400502291073863,2.076619307972976),linewidth(4pt) + dotstyle); label("$O_A$", (0.397469248676736,2.190423717864165), NE * labelscalefactor); dot((2.405158160631204,1.4419935792728382),linewidth(4pt) + dotstyle); label("$O_C$", (2.462078125673954,1.5487209587974664), NE * labelscalefactor); dot((-2.931677849389184,3.0820499608844165),linewidth(4pt) + dotstyle); label("$O_B$", (-2.880794846555333,3.194828036403346), NE * labelscalefactor); dot((0.3679210942093265,5.207446487757588),linewidth(4pt) + dotstyle); label("$Z$", (0.4253693686361579,5.315237153319394), NE * labelscalefactor); dot((-1.2684880323360295,4.15337345235106),linewidth(4pt) + dotstyle); label("$X$", (-1.2067876489900211,4.268982654841081), NE * labelscalefactor); dot((0.3479174057715809,2.960365486584159),linewidth(4pt) + dotstyle); label("$Y$", (0.397469248676736,3.0692774965859484), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$

Claim 1. $O_A, O_B, O_C$ are collinear.
Proof. $OP$ is the common radical axis of $\odot(AOP), \odot(BOP),\odot(COP)$ . $\blacksquare$

Claim 2. $XYZ$ and $ABC$ are similar.
Proof. Note that $\measuredangle XYZ=\measuredangle ABC$ because of the cyclic quadrilateral formed by the lines $AB,AC,\ell_2,\ell_3$ , and similarly the other equalities hold, and so $XYZ$ and $ABC$ are similar, as claimed. $\blacksquare$

Claim 3. $P,X,O_B,O_C$ are concyclic.
Proof. We know that $\measuredangle O_BXO_C=180-\measuredangle A$ from claim 2. We want to prove $\measuredangle O_BXO_C=\measuredangle O_BPO_C$ . This follows from:
$\begin{align*} \measuredangle O_BPO_C &= \measuredangle O_BPO+\measuredangle O_CPO \\ &= \measuredangle O_BOP+\measuredangle O_COP \\ &= \measuredangle O_COC+\measuredangle O_BOB \\ &= \frac{360-(\measuredangle OO_BB+\measuredangle OO_CC)}{2} \\ &= \frac{360-2(\measuredangle BPO+\measuredangle CPO)}{2} \\ &= \frac{360-2\measuredangle BPC}{2} \\ &= 180-\measuredangle A \end{align*}$ as claimed. $\blacksquare$

Claim 4. $P$ lies on the circumcircle of triangle $XYZ$ .
Proof. We present two proofs, one purely synthetic and the other complex bash.

First Proof. (Mohammed Imran) Consider the complete quadrilateral $XYZO_AO_BO_C$ . Since $P$ lies on the circumcircles of triangles $XO_CO_B$ and $ZO_AO_B$ , by Miquel we are done. $\blacksquare$

Second Proof. (Mohammed Imran [Synthetic]; Malay Mahajan [Complex Bash]) We claim that proving $X,P,A$ and $Y,P,B$ and $Z,P,C$ collinear proves this claim. That is because $\measuredangle ZPY=180-\measuredangle CPY=180-\measuredangle CPB=180-\measuredangle CAB=180-\measuredangle A=180-\measuredangle ZXY$ And we are done. So now it remains to prove $Y,P,B$ collinear, if we prove this we are done too, because the other arguments symmetrically follow.

We let $a$ be the complex number representing $A$ and similarly for other points. Without loss of generality, let the $(ABC)$ be unit circle (and hence $\overline{a} =a$ similarly for $b,c$ ), centered at origin, and trivially $o=0$ . Also we can assume without loss of generality, that $p=1$
Also, as $o_a$ is circumcenter of triangle with vertices $0,1,a$ , by the Complex Circumcenter lemma (EGMO 6.2.4) we get that $o_a=\frac{a}{a+1}$ Similarly, $o_b=\frac{b}{b+1},~o_c=\frac{c}{c+1}$ due to symmetry.
Note that line $O_AY$ is perpendicular to $BC$ and hence, has slope $+bc$ . Similarly $O_CY$ has slope $+ab$
Therefore , equation of $O_AY$ is $z-\frac{a}{a+1} = bc \left( \overline{z} - \overline{\left(\frac{a}{a+1}\right)}\right)$ Similarly, equation of $O_CY$ is $z-\frac{c}{c+1} = ab \left(\overline{z} - \overline{\left(\frac{c}{c+1}\right)}\right)$ Also, as $O_AY$ and $O_CY$ intersect at $Y$ , upon solving, we get that their intersection point $Y$ is $y= \frac{a+b+ab+abc}{(1+a)(1+b)}$
Now, to prove that $P(1),Y(y),B(b)$ collinear , it suffices to prove that $\frac{1-b}{1-y} = \overline{\left(\frac{1-b}{1-y}\right)}$ which is an easy computation, therefore $Y,P,B$ are collinear, as claimed. $\blacksquare$

Claim 5. $\odot(XYZ)$ is tangent to line $OP$ at $P$ .
Proof. Firstly, note that $\measuredangle PZY = \measuredangle PXY = \measuredangle PXO_C$ So it sufices to show that $\measuredangle PXO_C=\measuredangle PZY$ . This follows because,
$\begin{align*} \measuredangle PZY &= \measuredangle CZY \\ &= 90-\measuredangle PCB \\ &= \measuredangle PAB-90 \\ &= 90-\measuredangle XAB \\ &= PXO_C \end{align*}$
where the last step follows because $AB\perp XY$ , as claimed. $\blacksquare$

Remarks

This post has been edited 2 times. Last edited by rama1728, Feb 10, 2022, 5:32 PM
Reason: diag attach

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MathLuis

1517 posts

MathLuis

#36 h Feb 11, 2022, 12:52 AM

Y by

Point labeling explanation:
The triangle $E,F,G$ is formed by the intersections of $\ell_A,\ell_B,\ell_C$ and u might noted that the points $I,J,K$ are the foot of the perpendiculars from $O_A,O_B,O_C$ and the $H$ was a point that at the end i didnt used it xd. Also note that the location of the point $P$ is important here but since most of the things that we get w.r.t a vertex of $\triangle ABC$ are symetric there shouldn't be a problem.
Claim 1: $PEFG$ is cyclic
Proof: First note that $O_A,O_B,O_C$ are colinear by the radax and that $P$ is symetric to $O$ w.r.t. that line hence taking this in count and using circumcenter propeties and the cyclics formed by the perpendiculars we do angle chasing
$\angle O_BPO_A=\angle O_BOO_A=\angle O_BOP-\angle O_AOP=\angle PCB=\angle PCA=\angle ACB=\angle O_BFO_A$ $\angle O_BPO_C=\angle O_BOO_C=\angle O_BOP+\angle O_COP=180-\angle BPC=180-\angle BAC=\angle O_BEO_C$ Those angle chase tell that $PFO_AO_B, PEO_CO_B$ are cyclic hence $P$ is the miquel point of $EFO_CO_A$ and this completes the claim.
Claim 2: $P,F,C$ and $G,P,B$ colinear (for other cases u can use symetry to get the other colinearity, u need 2 colinearities)
Proof: First we prove a small colinearity with angle chase
$\angle IBG=\angle ACP=\angle JKG \implies I,K,J \; \text{colinear}$ Now by even more angle chasing using the cyclics and the colinearity
$\angle PFG=\angle PEG=\angle KJC=\angle O_AFC \implies P,F,C \; \text{colinear}$ $\angle GPF+\angle BPF=\angle GPF+\angle BAJ=\angle GPF+\angle GEF=180 \implies G,P,B \; \text{colinear}$ Finishing: We will end with angle chase using the colinearities and cyclics (said this a lot of times but yeah its true lol, this problem was interesting as i got some useless claims on the path to realice that i only needed some of them which is annoying but i will take it as this is a G7 that i solved so everything cool i guess)
$\angle PGF=90-\angle PBK=\angle PCO=\angle OPF \implies OP \; \text{tangent to} \; (EFG)$ Thus we are done :blush:

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JAnatolGT_00

559 posts

JAnatolGT_00

#37 h Jul 11, 2022, 4:39 PM • 2 Y

Y by PRMOisTheHardestExam, MS_asdfgzxcvb

Let $A'=\ell_B \cap \ell_C$ and define $B',C'$ analogously. Denote by $\ell$ tangent to $\Omega$ at antipode $Q$ of $P.$ Note that there exist spiral similarity $\varphi :$ $ABC\mapsto A'B'C'$ with rotation angle $90^\circ.$ $\measuredangle O_APO_B=\measuredangle O_BOO_A=\frac{1}{2} \measuredangle BOA=\measuredangle BCA=\measuredangle O_AC'O_B\implies P\in \odot (O_AO_BC').$ By all similar arguments we deduce that $P$ is the Miquel point of $\ell_A,\ell_B,$ $\ell_C,$ $\overline{O_AO_BO_C}.$ Next we observe that $\measuredangle PA'B'=\measuredangle PO_BO_A=\measuredangle O_AO_BO=\measuredangle PBO=\measuredangle QAB\implies \varphi (Q)=P.$ Since $\ell \perp OP$ we get $\varphi (\ell)=OP.$ The conclusion follows immediately.

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popop614

271 posts

popop614

#39 h Sep 16, 2023, 3:53 AM • 1 Y

Y by MS_asdfgzxcvb

+1 diagram, eepy geo goes hard

Here's the solution now.

Let $D$ be the intersection of line $PA$ with $(PO_BO_C)$ and define $E$ and $F$ similarly.
$\textbf{Claim.}$ $\triangle DEF$ is the aforementioned triangle.
$\emph{Proof.}$ We have
$\providecommand{\dang}{\measuredangle} \dang PDO_B = \dang PO_CO_B = 90^\circ - \dang OPO_C = \dang PCO = 90^\circ - \dang CAP,$ which establishes $O_BD\perp AC$ . Similarly, $O_CD \perp AB$ , and so on, so we know that $DEF$ is our desired triangle. Note we obtain the relation $\measuredangle PDO_B = \measuredangle PCO$ . I will call this $(\ast)$ . $\square$

As a corollary, we find $\overline{E-D-O_C}$ , $\overline{E-F-O_A}$ , and $\overline{D-F-OB}$ .

$\textbf{Claim.}$ $PDEF$ is cyclic.
$\emph{Proof.}$ Follows from angle chasing:
$\providecommand{\dang}{\measuredangle} \dang PDE = \dang PDO_C = \dang PO_BO_C = \dang PO_BO_A = \dang PFO_A = \dang PFE,$ yay. $\square$

It remains to prove that $OP$ is tangent to $(PDEF)$ . But just compute
$\providecommand{\dang}{\measuredangle} \dang PDF = \dang PDO_B \overset{(\ast)}{=} \dang PCO = \dang OPC = \dang OPD,$ as desired.
$[asy] unitsize(160); import olympiad; import math; import graph; import cse5; pair A,B,C,P,O,O_A,O_B,O_C,D,E,F; A = dir(110); B = dir(210); C = dir(-30); P = dir(80); O=circumcenter(A,B,C); O_A=circumcenter(P,O,A); O_B=circumcenter(P,O,B); O_C=circumcenter(P,O,C); path PBC,PCA,PAB; PBC = circumcircle(P,O_B,O_C); PCA = circumcircle(P,O_C,O_A); PAB = circumcircle(P,O_A,O_B); pair [] Ainte=intersectionpoints(PBC,L(P,A,0,10)); pair [] Binte=intersectionpoints(PCA,L(P,B,0,1000)); pair [] Cinte=intersectionpoints(PAB,L(C,P,0,1000)); D = Ainte[1]; E = Binte[1]; F = Cinte[1]; currentpen=rgb(0.99,0.92,0.98); fill(PBC); fill(PCA); fill(PAB); currentpen=rgb(1,0.9,0.9); fill(circumcircle(D,E,F)); currentpen=black; draw(A--B--C--cycle); draw(circumcircle(A,B,C)); currentpen = linetype("5 5")+rgb(0.6, 0.6, 0.6); draw(circumcircle(P,O,A)); draw(circumcircle(P,O,B)); draw(circumcircle(P,O,C)); currentpen = linetype("4 4")+orange+linewidth(1); draw(P--D); draw(P--B); draw(C--F); currentpen = black; draw(O_C--O_B); currentpen = black + linewidth(1) + linetype(""); currentpen = red; draw(O_B--F); draw(O_A--F); draw(D--O_C); currentpen=orange; draw(circumcircle(D,E,F)); currentpen=rgb(0.9,0.2,0.8); draw(PBC); draw(PCA); draw(PAB); currentpen=black; dot(A); dot(B); dot(C); dot(O); dot(O_A); dot(O_B); dot(O_C); dot(P); dot(D); dot(E); dot(F); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$O$",O,ESE); label("$P$",P,NE); label("$D$",D,NW); label("$E$",E,dir(120)*2); label("$F$",F,NE); label("$O_A$",O_A,dir(250)*1.5); label("$O_B$",O_B,W); label("$O_C$",O_C,right); [/asy]$

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math_comb01

662 posts

math_comb01

#40 h Dec 24, 2023, 10:38 AM

Y by

This felt easiest in the shortlist lol
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(12.5 cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 1.8366426566663001, xmax = 10.91942116823776, ymin = -6.306120037901597, ymax = 0.8055118964383086; /* image dimensions */ pen zzttff = rgb(0.6,0.2,1); pen ccqqqq = rgb(0.8,0,0); draw((4.022126346235072,0.4247610767458665)--(3.382966625751293,-1.8245894780335803)--(10.512055815762658,-2.463749198517359)--cycle, linewidth(0) + zzttff); /* draw figures */ draw(circle((6.993165067059032,-1.6349533602909858), 3.6151756033107594), linewidth(0.4) + blue); draw(circle((4.812858577594672,-1.6072930609311176), 2.180481937591111), linewidth(0.4) + ccqqqq); draw(circle((5.213713324169409,-2.2180353920286784), 1.8725472386586859), linewidth(0.4) + ccqqqq); draw(circle((7.7888082498901605,-6.141450272782654), 4.576195198708485), linewidth(0.4) + ccqqqq); draw((xmin, 11.153846153846134*xmin-55.28917719564082)--(xmax, 11.153846153846134*xmax-55.28917719564082), linewidth(0.4)); /* line */ draw((xmin, 2.246808510638297*xmin-13.932250860800794)--(xmax, 2.246808510638297*xmax-13.932250860800794), linewidth(0.4)); /* line */ draw((xmin, -0.28415300546448147*xmin-3.9282369995898163)--(xmax, -0.28415300546448147*xmax-3.9282369995898163), linewidth(0.4)); /* line */ draw(circle((4.445044063230293,-4.3411457121527), 0.8642355307754483), linewidth(0.4) + ccqqqq); draw((xmin, 0.6563402046397618*xmin-6.224848751484146)--(xmax, 0.6563402046397618*xmax-6.224848751484146), linewidth(0.4)); /* line */ draw((4.812858577594672,-1.6072930609311176)--(7.7888082498901605,-6.141450272782654), linewidth(0.4)); draw((3.382966625751293,-1.8245894780335803)--(4.490378035588752,-5.204191414074056), linewidth(0.4)); draw((3.9526534866540928,-5.051395367382239)--(4.022126346235072,0.4247610767458665), linewidth(0.4)); draw((3.9708301182016617,-3.6186332991139367)--(10.512055815762658,-2.463749198517359), linewidth(0.4)); draw((xmin, 3.5192307692307616*xmin-13.730029718658297)--(xmax, 3.5192307692307616*xmax-13.730029718658297), linewidth(0.4)); /* line */ draw((xmin, -0.44507575757575774*xmin + 2.214912007361855)--(xmax, -0.44507575757575774*xmax + 2.214912007361855), linewidth(0.4)); /* line */ draw((xmin, -0.08965517241379327*xmin-1.5212890219317399)--(xmax, -0.08965517241379327*xmax-1.5212890219317399), linewidth(0.4)); /* line */ draw(circle((6.730821017475573,-3.5921535150301054), 2.7601179219446292), linewidth(0.4) + ccqqqq); draw((4.812858577594672,-1.6072930609311176)--(8.698519652908601,-1.6565882169440589), linewidth(0.4)); draw((6.851875076301115,-0.834691483037318)--(7.7888082498901605,-6.141450272782654), linewidth(0.4)); /* dots and labels */ dot((4.022126346235072,0.4247610767458665),dotstyle); label("$A$", (4.065450454143153,0.528034927901925), NE * labelscalefactor); dot((3.382966625751293,-1.8245894780335803),dotstyle); label("$B$", (3.4282811189855296,-1.7226115946709646), NE * labelscalefactor); dot((10.512055815762658,-2.463749198517359),dotstyle); label("$C$", (10.550189978086056,-2.3597809298285863), NE * labelscalefactor); dot((3.9708301182016617,-3.6186332991139367),dotstyle); label("$P$", (4.014065830340118,-3.5107965030165476), NE * labelscalefactor); dot((6.993165067059032,-1.6349533602909858),linewidth(4pt) + dotstyle); label("$O$", (7.035481709958524,-1.547903873740649), NE * labelscalefactor); dot((4.812858577594672,-1.6072930609311176),linewidth(4pt) + dotstyle); label("$O_A$", (4.8567736607098775,-1.5273500242194353), NE * labelscalefactor); dot((5.213713324169409,-2.2180353920286784),linewidth(4pt) + dotstyle); label("$O_B$", (5.257573726373543,-2.1336885850952365), NE * labelscalefactor); dot((7.7888082498901605,-6.141450272782654),linewidth(4pt) + dotstyle); label("$O_C$", (7.826804916525249,-6.059473843647035), NE * labelscalefactor); dot((3.9526534866540928,-5.051395367382239),linewidth(4pt) + dotstyle); label("$D$", (3.9935119808189046,-4.970119819022713), NE * labelscalefactor); dot((4.490378035588752,-5.204191414074056),linewidth(4pt) + dotstyle); label("$E$", (4.527912068370459,-5.124273690431815), NE * labelscalefactor); dot((4.643174082280568,-3.4999278163576473),linewidth(4pt) + dotstyle); label("$F$", (4.682065939779561,-3.4183041801710865), NE * labelscalefactor); dot((8.698519652908601,-1.6565882169440589),linewidth(4pt) + dotstyle); label("$G$", (8.741451220219256,-1.5787346480224693), NE * labelscalefactor); dot((6.851875076301115,-0.834691483037318),linewidth(4pt) + dotstyle); label("$H$", (6.891604763310028,-0.7565806671739252), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]$
Let $H = OO_C \cap AC$ let $G = OO_A \cap AC$
Claim 1: $O_AO_CGHPE$ is a cyclic hexagon
Proof
Claim 2: $B-P-E$
Proof
By Claim 2, $PFDE$ is cyclic, which implies $OP$ tangent as required after some angle chase.

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HamstPan38825

8857 posts

HamstPan38825

#41 h Mar 13, 2024, 3:46 AM

Y by

Let $DEF$ be the formed triangle. The main claim is the following:

Claim. $D, A, P$ are collinear.

Proof. Complex numbers. We will let $D =\overline{AP} \cap \ell_C$ and show that the expression for $d$ is symmetric in $b$ and $c$ . First, let $f$ be the foot from $O_C$ to $\overline{AB}$ . We have the following formulas:
$\begin{align*} o_C &= \frac{cp(\overline c-\overline p)}{\overline cp-c\overline p} = \frac{cp}{c+p} \\ f &= \frac 12\left(a+b+\frac{cp-ab}{c+p}\right) \end{align*}$ using complex circumcenter and foot formula. Note that their conjugates are
$\begin{align*} \overline o_c &= \frac 1{p+c} \\ \overline f &= \frac 12\left(\frac 1a+\frac 1b+\frac 1{p+c}-\frac{cp}{ab(p+c)} \right). \end{align*}$ To compute $d$ , we first set up the following preliminary calculations:
$\begin{align*} \overline o_c f - o_c \overline f &= \frac 12\left[\frac{(p+c)(a+b)+cp-ab}{(p+c)^2} - \frac{cp}{c+p} \left(\frac 1a+\frac 1b+\frac 1{p+c}-\frac{cp}{ab(p+c)}\right)\right] \\ &= \frac{(cp-ab)(ab-ac-ap-bc-bp+cp)}{ab(c+p)^2} = \frac{S(cp-ab)}{2ab(c+p)^2}. \\ o_c-f &= \frac 12\left(\frac{ab+cp}{c+p}-a-b\right) = \frac S{2(c+p)}. \\ \overline o_c-\overline f &= \frac 1{c+p}-\frac 12\left(\frac 1a+\frac 1b+\frac 1{c+p}-\frac{cp}{ab(p+c)}\right)= \frac S{2ab(c+p)}. \end{align*}$ Here we denote $S = ab-ac-ap-bc-bp+cp$ . Now, the complex intersection formula yields
$\begin{align*} d &= \frac{\frac{S(a-p)(cp-ab)}{2ab(c+p)^2} - \frac S{2(c+p)} \cdot \left(\frac pa-\frac ap\right)}{\frac {S(a-p)}{2ab(c+p)} - \frac {S\left(\frac 1a-\frac 1p\right)}{2(c+p)}} \\ &= \frac{\frac{(a-p)(cp-ab)}{ab(c+p)^2} - \frac{p^2-a^2}{ap(c+p)}}{\frac{a-p}{ab(c+p)}-\frac{p-a}{ap(c+p)}} \\ &= \frac{\frac{cp-ab}{ab(c+p)^2} +\frac{a+p}{ap(c+p)}}{\frac 1{ab(c+p)} +\frac 1{ap(c+p)}} \\ &= \frac{p(cp-ab)+b(a+p)(c+p)}{p(c+p)+b(c+p)} \\ &= \frac{(b+c)p^2+(a+p)bc}{(c+p)(b+p)}. \end{align*}$ This is symmetric in $b$ and $c$ , as required. $\blacksquare$

From here, the problem succumbs to just angle chasing. First, observe that $\triangle DEF \sim \triangle ABC$ as corresponding sides are perpendicular. Then $\measuredangle EPF = \measuredangle BAC = \measuredangle EDF$ , hence $EDFP$ is cyclic. To finish, notice that $\angle DFP = \angle DEB = 90^\circ - \angle ABP = \angle APO$ , hence $\overline{OP}$ is tangent to $(DEF)$ at $P$ , as required.

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bin_sherlo

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bin_sherlo

#42 h Oct 26, 2024, 7:08 PM

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Mine is similar. :dry:

Let $l_c\cap l_b=D,l_c\cap l_a=E,l_b\cap l_a=F$ and $l_a\cap BC=X,l_b\cap AC=Y,l_c\cap AB=Z$ .
Claim: $A,D,P$ are collinear.
Proof
Let $M$ be the miquel point of $BCYZ$ . Since $M$ lies on both $(PDE),(PBA)$ , we conclude that $M$ is the miquel point of $BPDZ$ . Similarily $M$ is the miquel point of $PCYD$ . Hence $M\in (PDE),(PDF)$ which implies $M,D,E,F,P$ are concyclic. Let $A'$ be the antipode of $A$ on $(ABC)$ . Since $\measuredangle AMD=90=\measuredangle AMA'$ so $M,D,A'$ are collinear. $\measuredangle DMP=\measuredangle CMP-\measuredangle CMA'=\measuredangle CAP-\measuredangle CAA'=\measuredangle OAP=\measuredangle DPO$ as desired. $\blacksquare$

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OronSH

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OronSH

#43 h Dec 23, 2024, 2:47 AM

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Move $P$ with degree $2$ on $\Omega$ .

Now $AP$ has degree $1$ , so its pole does, so its midpoint with $O$ , which is $O_A$ , does as well. Similarly $O_B$ and $O_C$ also have degree $1$ . This then implies $D,E,F$ each have degree $2$ .

Consider the three cases when $P$ is a vertex of the circumtangential triangle. Let $\ell$ be the tangent to $\Omega$ at $P$ . Then $\measuredangle PAO=\tfrac12\measuredangle PO_AO=\measuredangle(O_AP,\ell)$ implies $AO,PO_A$ isogonal in $\measuredangle(AP,\ell)$ . Then they are also isogonal in $\measuredangle BAC$ , implying $PO_A\perp BC$ . Similarly $PO_B\perp AC,PO_C\perp AB$ so $D=E=F=P$ .

By Zack's lemma, lines $PD,PE,PF$ have degree $4-3=1$ so they pass through fixed points.

When $P=A$ clearly $A\in PD$ and when $P=2O-A$ then $D$ is the orthocenter of $\triangle OO_BO_C$ so $A\in PD$ again. Thus $A,P,D$ are always collinear. Similarly $B,P,E$ and $C,P,F$ are collinear.

Since $\triangle ABC$ and $\triangle DEF$ are paralogic, it follows their circumcircles are orthogonal, implying the result.

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TestX01

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TestX01

#44 h Jan 22, 2025, 3:40 AM

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Let $\triangle XYZ$ be the triangle formed by $\ell_A$ , $\ell_B$ , $\ell_C$ . We will prove that $A,X,P$ are collinear, and then show that $(XYZ)$ is tangent to $OP$ at $P$ .

Let us complex bash, and use lowercase to denote complex numbers.

Let $(ABC)$ be the unit circle, and rotate the diagram such that $p=1$ . Using the circumcentre formula, we derive that
$o_B=\frac{b(\overline{b}-1)}{\overline{b}-b}=\frac{1-b}{\frac{1}{b}-b}=\frac{b}{b+1}$ Similarly, $o_C=\frac{c}{c+1}$ . By perpendicularity, $X$ , $O_B$ have the same foot onto $AC$ , and $X$ , $O_C$ have the same foot onto $AB$ .

Hence, $\frac{1}{2}(a+b+x-\overline{x}ab)=\frac{1}{2}\left(a+b+\frac{c}{c+1}-\frac{ab}{c+1}\right)$ noting $\frac{\frac{1}{c}}{\frac{1}{c}+1}=\frac{1}{c+1}$ .

Hence, $x-\overline{x}ab=\frac{c-ab}{c+1}$ . Analogously, we also have $x-\overline{x}ac=\frac{b-ac}{b+1}$ . Hence,
$\overline{x}(ac-ab)=\frac{c-ab}{c+1}-\frac{b-ac}{b+1}=\frac{a(c^2-b^2)+a(c-b)+(c-b)}{(b+1)(c+1)}$ Thus,
$\overline{x}=\frac{c+b+\frac{1}{a}+1}{(b+1)(c+1)}$ Taking the conjugate,
$x=\frac{\frac{1}{c}+\frac{1}{b}+a+1}{\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right)}=\frac{b+c+abc+bc}{(b+1)(c+1)}$ However, we need to verify $x$ actually satisfies one of the equations (the other follows by symmetry).
$x-\overline{x}ab=\frac{b+c+abc+bc-abc-ab^2-ab-b}{(b+1)(c+1)}=\frac{(b+1)(c-ab)}{(b+1)(c+1)}=\frac{c-ab}{c+1}$ As desired.

Now, let's check that $A$ , $X$ , $P$ are collinear. Indeed, we just need $a+1=a\overline{x}+x$ by chord formula.
$\frac{a(b+c)+1+a+b+c+abc+bc}{(b+1)(c+1)}=\frac{(a+1)(b+1)(c+1)}{(b+1)(c+1)}=a+1$ As desired.

Analogously, $C,Z,P$ and $B,Y,P$ are collinear.

Now, to finish,
$\measuredangle PZX=90^\circ-\measuredangle ACP=\measuredangle OPA=\measuredangle OPX$ From our collinearities.

Thus, $(ZPX)$ is tangent to $OP$ . Analogously, $(XPY)$ is also tangent to $OP$ , hence $(XYZP)$ is tangent to $OP$ .

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