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Tentang KSN Matematika SMA 2022 Esai P4

Published on 03 September 2022

Setelah sekian lama hiatus, finally I'm back! Let me start with a problem I proposed to KSN-P (Kompetisi Sains Nasional tingkat Provinsi) Matematika this year. It's a geometry problem, which was inspired by a geometry problem on AoPS. Here's the statement, in English and Indonesian, respectively.

Problem Statement

English Statement

Suppose $ABC$ is a triangle with circumcenter $O$. Point $D$ is the reflection of $A$ with respect to $BC$. Suppose $\ell$ is the line which is parallel to $BC$ and passes through $O$. The line through $B$ and parallel to $CD$ meets $\ell$ at $B_1$. Lines $CB_1$ and $BD$ intersect at point $B_2$. The line through $C$ parallel to $BD$ and $\ell$ meet at $C_1$. Finally, $BC_1$ and $CD$ intersects at point $C_2$. Prove that points $A, B_2, C_2, D$ lie on a circle.

Indonesian Statement

Diberikan segitiga $ABC$ dengan titik pusat lingkaran luar $O$. Titik $D$ merupakan refleksi titik $A$ terhadap $BC$. Misalkan $\ell$ adalah garis yang sejajar dengan $BC$ dan melalui $O$. Garis melalui $B$ sejajar $CD$ dan $\ell$ bertemu pada titik $B_1$. $CB_1$ dan $BD$ berpotongan pada titik $B_2$. Garis melalui $C$ sejajar $BD$ dan $\ell$ bertemu pada titik $C_1$. $BC_1$ dan $CD$ berpotongan pada titik $C_2$. Buktikan bahwa $A, B_2, C_2, D$ terletak pada satu lingkaran.

Official Solutions (In Indonesian)

Well, here are the two solutions that I submitted in the proposal. Both of them basically prove a common claim, which is stated below.

The common claim used in both solutions

Perhatikan bahwa apabila dapat dibuktikan bahwa $C_2A \perp AB$, maka dengan cara yang sama dapat dibuktikan bahwa $B_2A \perp AC$, sehingga: $$\angle B_2AC_2 = \angle B_2AC + \angle BAC_2 - \angle BAC = 90^{\circ} + 90^{\circ} - \angle BAC = 180^{\circ} - \angle B_2DC_2, $$ yang mengakibatkan $A, B_2, C_2, D$ terletak pada satu lingkaran. Sehingga cukup dibuktikan bahwa $C_2A \perp AB$.

We have two ways to prove this claim. Well, in fact I can just say that there is only a solution with two ways to finish, but well whatever :)

Solusi 1

Misal $AO$ memotong $BC$ di $E$, dan $X, Y$ adalah dua titik sehingga segiempat $OECC_1$ dan $XYDB$ homotetik. Misal garis tegak lurus $AB$ melalui $A$ memotong $DC$ pada $F$. Perhatikan bahwa $\angle DAF = 90^{\circ} - \angle BAD = \angle ABE, \angle ADF = \angle BAE$, sehingga segitiga $ABE$ dan $DAF$ sebangun. Selain itu, $O$ merupakan perpotongan garis sumbu $AB$ dengan garis $AE$, dan $C$ merupakan perpotongan garis sumbu $AD$ dengan garis $FD$. Maka, $$ \frac{FC}{FD} = \frac{EO}{EA} = \frac{EO}{XY} = \frac{CC_1}{DB}. $$ Sehingga, dari Thales, $B, C_1, F$ segaris. Akibatnya, $F \equiv C_2$, sehingga $AC_2 \equiv AF \perp AB$. $\blacksquare$

Solusi 2

Misal $Y$ adalah titik sehingga $AYCB$ adalah trapesium sama kaki dengan $AY \parallel BC$. Mudah dilihat bahwa $CY \parallel BD$. Misalkan pula $AO$ bertemu $BC$ pada $E$. Akibatnya, \[ \frac{C_2C}{C_2D} = \frac{CC_1}{BD} = \frac{CC_1}{CY} = \frac{EO}{EA}. \] Karena $ABO \sim ADC$, maka persamaan di atas menyebabkan $ABE \sim DAC_2$. Akibatnya, $\angle C_2AB = \angle C_2AD + \angle DAB = \angle B + 90^{\circ} - \angle B = 90^{\circ}$. $\blacksquare$

Authorship Note

I found this configuration while trying to solve a problem from AoPS (unfortunately I don't have the link to the thread now, need to dig my bookmark first 😕). At first, the problem only asked to prove that $B_2A \perp AC$. The points $C_1$ and $C_2$ weren't even drawn, so the problem was kinda asymmetric. But then, I realized that if $C_2$ was also drawn, then $A, B_2, C_2, D$ would be concyclic. I thought that this version would be easier than the first one, because there might be another approach that doesn't rely on the fact that $B_2A \perp AC$. I was planning to propose this to the national stage of KSN, but Mr.bermatematika.com asked me if I have a problem ready for KSN-P, so why don't we just try submitting this one, right? Well, if it isn't chosen for the KSN-P, I've asked him to put this for KSN national proposal.

IndoMathXdz pointed out on AoPS that this problem could be generalized by letting $\ell$ as any line passing through $O$. Damn, I should've realized this and maybe try proposing this one for APMO, or maybe even IMO, idk haha. Well, the cat is out of the box now, and the only thing I could do is trying to solve the general version, which luckily I can do, and whose solution I provided below. Enjoy!

Solusi for General Version

Define $E = BB_1 \cap CC_1$ and $M$ as the midpoint of $BC$. Note that \[ \frac{DB}{BB_2} = \frac{CE}{BB_2} = \frac{EB_1}{BB_1} \implies \frac{DB}{DB_2} = \frac{EB_1}{EB}. \] Similarly, we have $DC/DC_2 = EC_1/EC$. Now let $O', B_3, C_3$ be the reflection of $O, B_1, C_1$ with respect to $M$ respectively. Obviously, $O'$ is the circumcenter of $\triangle DBC$. We have: \[ \frac{DB_3}{DC} = \frac{EB_1}{EB} = \frac{DB}{DB_2} \implies DB_3 \times DB_2 = DB \times DC. \] Similarly, $DC_3 \times DC_2 = DB \times DC$. Therefore, the composition of inversion with center $D$ and radius $\sqrt{DB \times DC}$ followed by reflection w.r.t. $\angle BDC$ will map $B \mapsto C$, $B_3 \mapsto B_2$, and $C_3 \mapsto C_2$. It's known by the property of $\sqrt{bc}$ inversion that $O' \mapsto A$. Since $B_3, O', C_3$ are collinear, therefore $D, B_2, A, C_2$ are concyclic, as desired. $\square$

Thanks for reading! Have a nice day! 😄