Dealing with Multiple Circles

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Source: PEMNAS Brawijaya University Senior High School Semifinal 2023 P4

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#1 h Dec 1, 2024, 2:43 PM

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Given non-isosceles triangle $ABC$ and $\ell$ is tangent line to the circumcircle of triangle $ABC$ through point $C$ . Points $D$ and $E$ are the midpoints of segments $BC$ and $CA$ respectively, then line $AD$ and line $BE$ intersect $\ell$ at points $A_1$ and $B_1$ respectively. Line $AB_1$ and line $BA_1$ intersect the circumcircle of triangle $ABC$ at points $X$ and $Y$ respectively. Prove that $X$ , $Y$ , $D$ and $E$ concyclic.

This post has been edited 2 times. Last edited by Wildabandon, Mar 22, 2025, 7:56 PM
Reason: Wording

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#2 h Dec 2, 2024, 1:19 AM • 1 Y

Y by MS_asdfgzxcvb

When in doubt, just invert

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First let $(CDE) \cap AD=L$ and $(CDE) \cap BE=K$ , then from Reim's theorem $BKLC$ is cyclic, now consider inversion with center $C$ and some arbitrary radius, denote $(\cdot)'$ to be the inverse of $(\cdot)$ .
Notice that $\ell$ is fixed and from anti-parallels $B'C' \parallel D'E' \parallel \ell$ , also we have that $(CA'D') \cap \ell=A_1'$ and $(CB'E') \cap \ell=B_1'$ , then we have $(CA'B_1') \cap A'B'=X'$ and $(CB'A_1') \cap A'B'=Y'$ , finally from double Reim's we can get a new definition for $K',L'$ which is $B_1'X' \cap D'E'=K'$ and $A_1'Y'\cap D'E'=L'$ .
From the inversion we get that $K'B'C'L'$ is cyclic but also that from the consequence of the reim's and the parallel lines we get $E'X'A'K', D'Y'B'L'$ are cyclic so yet by another reim's $X'E' \parallel B'L'$ and then by yet another Reim's we have $E'X'Y'D'$ is an isosceles trapezoid (because of the parallel lines) which is enough to conclude by inverting back, thus we are done :cool:

:cool:

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iStud

#3 h Dec 2, 2024, 3:25 AM

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Wildabandon wrote:

A non-isosceles triangle $ABC$ and $\ell$ is tangent to the circumcircle of triangle $ABC$ through point $C$ . Points $D$ and $E$ are the midpoints of segments $BC$ and $CA$ respectively, then line $AD$ and line $BE$ intersect $\ell$ at points $A_1$ and $B_1$ respectively. Line $AB_1$ and line $BA_1$ intersect the circumcircle of triangle $ABC$ at points $X$ and $Y$ respectively. Prove that $X$ , $Y$ , $D$ and $E$ concyclic.

Let $BE$ and $AD$ meet $(ABC)$ for the second time at $P$ and $Q$ . Notice that since $\angle{EPQ}=\angle{BPQ}=\angle{BAQ}=\angle{BAD}=\angle{ADE}=180^\circ-\angle{EDQ}$ , then $PQDE$ is cyclic. Construct $(CDE)$ and note that by Radax Theorem within $(ABC),(CDE),(PQDE)$ , the tangent to $(ABC)$ at $C$ , $PQ$ , and $DE$ meet at one point, let it be $Z$ . Next, Pascal Theorem at $XAQPBY$ gives us $XA\cap PB=B_1$ , $AQ\cap BY=A_1$ , and $QP\cap YX=Z'$ are collinear. Thus, $Z'=PQ\cap A_1B_1$ . But $Z$ is also defined similarly, so we have $Z=Z'$ . For the final act, simply see that $ZD\cdot ZE=ZC^2=ZY\cdot ZX$ , thus $XYDE$ is cyclic by the converse of PoP Theorem. Done. $\blacksquare$

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This post has been edited 3 times. Last edited by iStud, Dec 3, 2024, 12:01 AM
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#4 h Mar 22, 2025, 3:56 PM • 1 Y

Y by MS_asdfgzxcvb

Another solution using inversion, but I only provide a sketch.

Let $P:= AA_1\cap (ABC)$ and $Q:= BB_1\cap (ABC)$ , take inversion with centre $C$ (arbitrary radius). We have
- $P',Q',X',Y',A',B'$ collinear
- $C,A_1',P',D',A'$ concyclic.
- $C,B_1',Q',E',B'$ concyclic.
- $B_1',C,A',X'$ concyclic.
- $A_1',C,B',Y'$ concyclic.
- $P'D'Q'E'$ cyclic and isosceles trapezoid since $D,E,P,Q$ concyclic.
- $\ell \parallel A'B'\parallel D'E'$ .

It's sufficient to show that $D'Y'X'E'$ isosceles trapezoid iff $X'P'=Y'Q'$ by some angle chasing (hence $\triangle P'Y'D'\cong \triangle Q'X'A'$ ). By angle chasing, we have $A_1'B_1'X'P'$ and $Q'B_1'A_1'Y'$ parallelogram and we are done.

Remark. I am the author of the problem.

This post has been edited 6 times. Last edited by Wildabandon, Mar 22, 2025, 7:18 PM
Reason: typo and add some detail

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#5 h Mar 22, 2025, 5:53 PM

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Complex bash:

Consider $P$ and $Q$ as @above.

Compute $p=bc\cdot\frac{b+c-2a}{2bc-ab-ac}$ , $A_1=AP\cap CC\implies a_1=\frac{c(3a^2b+b^2c+bc^2-a^2c-2ab^2-2abc)}{2a^2b+2bc^2-ab^2-ac^2-2abc}$ , and
$Y=BA_1\cap (ABC)\implies y=\frac{a^2c+ab^2+abc-bc^2-2a^2b}{ac^2+b^2c+abc-2bc^2-a^2b}$ .

We similarly get $x=\frac{b^2c+a^2b+abc-ac^2-2ab^2}{bc^2+a^2c+abc-2ac^2-ab^2}$ .

To prove that $X,Y,D,E$ are concyclic, we just have to prove that $\frac{(d-e)(y-x)}{(d-x)(y-e)}\in\mathbb{R}$ .

$d-e=\frac{b-a}{2}$

$d-x=\frac{(c-b)(ab^2+bc^2+ca^2-3abc)}{2(bc^2+a^2c+abc-2ac^2-ab^2)}$

$y-e=\frac{(a-c)(a^2b+b^2c+c^2a-3abc)}{2(ac^2+b^2c+abc-2bc^2-a^2b)}$

$y-x=\frac{c(a-b)(a+b)(a-c)^2(b-c)^2}{(bc^2+a^2c+abc-2ac^2-ab^2)(ac^2+b^2c+abc-2bc^2-a^2b)}$

So $\frac{(d-e)(y-x)}{(d-x)(y-e)}=2\cdot\frac{(a-b)^2(a+b)(a-c)(b-c)}{(a^2b+b^2c+c^2a-3abc)(ab^2+bc^2+ca^2-3abc)}$ , which can be seen that it's real by conjugating, so we're done.

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