Euclid cannot draw circles under the current conditions? I am utterly nonplussed.

RIP GEO MAINS

So nice Ok so here is my solution from contest First draw the incentre $I$ of $\triangle BIC$.Mark the lines $BI$ and $CI$.Let $\ell_{XYZ}$ denote the angle bisector of $\angle XYZ$. First draw $\ell_{BAI}$ and mark $CI \cap \ell_{BAI}=E$ .Clearly $E$ lies on the opposite side of $I$ w.r.t $C$ Now draw $\ell_{EIB}$ and mark $\ell_{EIB}\cap \ell_{ICB}=K_C$.Note that $K_C$ is the $C$-excentre of $\triangle BIC$.Similarly construct $K_B$ which is the $B$ excentrez of $\triangle BIC$. Now note that $K_BCBK_C$ is cyclic.So by fact $5$ we have that $\ell_{K_BBK_C} \cap \ell_{K_BCK_C}=M$ lie on the circle $(K_BCBK_C)$ and $M$ is also the midpoint of arc $K_BK_C$ (on which $B$,$C$ doesn't lie). Now since $\triangle MK_BK_C$ is isosceles, we have that $\ell_{K_BMK_C}$ is perpendicular bisector of $K_BK_C$ , since $K_BK_C$ is nothing other than $\ell_{EIB}$ , we can mark $\ell_{K_BMK_C}\cap K_BK_C=L$ .Now note that $L$ is the centre of $(K_BCBK_C)$ Thus we have $\ell_{BLC}$ is the perpendicular bisector of $BC$ This we have drawn the perpendicular bisector of $BC$.Repeat this process with $CA$ or $AB$ and mark the intersection of the perpendicular bisectors to get the circumcentre $\blacksquare$

Yea incenter-excenter lemma

Scilyse wrote: Euclid cannot draw circles under the current conditions? I am utterly nonplussed. Euclid can not draw circles but he can mark the centre. So may be we have to find the locus of all points satisfying a certain property and forming the circle

Aparently this trolled many people, so uhh lets solve ig. Draw internal angle bisectors of $(B,A,C), (C, B, A), (A, B, C)$ to get they meet at $I$ the incenter of $\triangle ABC$, then internal angle bisector of $(A,B,I)$ meets $CI$ (which is already drawn when we created the incenter), at $D$, noticed $D$ is always outside $\triangle BIC$ by definition so now we use device on $(D,I,B), (B,I,C)$ to draw internal and external angle bisectors of $\angle BIC$, now use device on $(I,B,C), (I,C,B)$ to draw $J_B, J_C$ which are the $B,C$-excenters of $\triangle BIC$, notice since $\angle J_BBJ_C=90=\angle J_CCJ_B$ we have that $J_BBCJ_C$ is cyclic with diameter $J_BJ_C$ and now using device on $(B, J_B, C), (C, J_C, B)$ to draw those two angle bisectors that meet at $M$, $M$ ends up by being midpoint of arc $BC$ on $(J_BJ_C)$ which means that using device on $(B,M,C)$ gives the perpendicular bisector of $BC$, now repeat cyclically to win .

This problem was proposed by Shankhadeep Ghosh.

S.Ragnork1729 wrote: Euclid has a tool called splitter which can only do the following two types of operations : • Given three non-collinear marked points $X,Y,Z$ it can draw the line which forms the interior angle bisector of $\angle{XYZ}$. • It can mark the intersection point of two previously drawn non-parallel lines . Suppose Euclid is only given three non-collinear marked points $A,B,C$ in the plane . Prove that Euclid can use the splitter several times to draw the circle passing through $A,B$ and $C$. I think this should be "draw the center of the circle passing through $A$,$B$, and $C$."

mathscrazy wrote: This problem was proposed by Shankhadeep Ghosh. Shankhadeep da orzzz

solved in contest

S.Ragnork1729 wrote: Euclid has a tool called splitter which can only do the following two types of operations : • Given three non-collinear marked points $X,Y,Z$ it can draw the line which forms the interior angle bisector of $\angle{XYZ}$. • It can mark the intersection point of two previously drawn non-parallel lines . Suppose Euclid is only given three non-collinear marked points $A,B,C$ in the plane . Prove that Euclid can use the splitter several times to draw the circle passing through $A,B$ and $C$. Proposed by Shankhadeep Ghosh The most beautiful problem from INMO 2025 You want to construct the circumcenter.

Solved using geogebra, probably same as others. At first look I thought it would be harder than INMO 2023 P6, but I was wrong

@below i was getting goosebumps after reading the problems. 2 combo and density vibes from p6. This problem felt nice

L13832 wrote: Solved using geogebra, probably same as others. At first look I thought it would be harder than INMO 2023 P6, but I was wrong

The test was alr harder than inmo 23 minus the geo and P1 how much more hardness do you want

HoRI_DA_GRe8 wrote: L13832 wrote: Solved using geogebra, probably same as others. At first look I thought it would be harder than INMO 2023 P6, but I was wrong

The test was alr harder than inmo 23 minus the geo and P1 how much more hardness do you want Cutoff should crash like stock market if it is harder than inmo 23 .

What should be the expected cutoff?

@Supercali yes, there is a typo here, in the original paper it is "construct the center of the circle passing through A, B and C"

S.Ragnork1729 wrote: HoRI_DA_GRe8 wrote: L13832 wrote: Solved using geogebra, probably same as others. At first look I thought it would be harder than INMO 2023 P6, but I was wrong

The test was alr harder than inmo 23 minus the geo and P1 how much more hardness do you want Cutoff should crash like stock market if it is harder than inmo 23 . I feel so too (However lets see cant rly say stuff)

@above During the exam, I also thought it was like inmo 2023 p6 and I remembered it had like 8 lemmas in the original question and then I messed up the rest of the paper....

My problem , proposed for the first time.

People came up with really cool solutions to this, I am happy . I am sharing the solution that I originally came up with.

Shankha013k wrote: My problem , proposed for the first time. People came up with really cool solutions to this, I am happy . I am sharing the solution that I originally came up with. Congrats! very cool problem, seems like my solution is the same as yours

InterLoop wrote: solved in contest

Can you confirm why does the intersection of angle bisector of BIC and AI give A-excenter of AIB?

Shankha013k wrote: Can you confirm why does the intersection of angle bisector of BIC and AI give A-excenter of AIB? Oh sorry, $R$ is the intersection of the angle bisector of $\angle IBC$ and $AI$ rather than $\angle BIC$ and $AI$. Analogous to your solution (where you called it $Y$ I think), the point of $R$ is just for the external view in triangle $\triangle AIB$ since after this we can construct the angle bisectors of $\angle BAI$ and $\angle BIR$, which leads to excenter of $AIB$. Edited!

Sol:- Clearly we can mark the incenter of any triangle. Let $I$ be the incenter of $ABC$. $I_A$ be the incenter of $BIC$. Let the bisector of $\angle BAI$ meet $BI,CI$ at $D,E$.The internal angle bisector of $DIE$ ,say $l$, is just the external angle bisector of $BIC$. $l \cap CI_A=I'$ is the $C$ excenter of $BIC$. Use the bisectors of $\angle I'BI,\angle I'I_AI$ to mark the arc midpoint of $II'$ (say $N$) in cyclic quad $BI_AI'I$. The bisector of $\angle I'NI$ meet $CI_A$ at $F$ i.e. the center of $(BI_AI'I)$. The bisector of $\angle BFI$ is just the perpendicular bisector of $BI$ ,say $m$ . In a similar way we draw the perpendicular bisector of $CI$ which would intersect $m$ at $M_A$ i.e. the arc midpoint of $BC$ in $(ABC)$. The bisector of $\angle BM_AC$ is the perpendicular bisector of $BC$. Similarly draw the perpendicular bisectors of other sides to get $O$ ,the circumcenter of $ABC$.

InterLoop wrote: Shankha013k wrote: Can you confirm why does the intersection of angle bisector of BIC and AI give A-excenter of AIB? Oh sorry, $R$ is the intersection of the angle bisector of $\angle IBC$ and $AI$ rather than $\angle BIC$ and $AI$. Analogous to your solution (where you called it $Y$ I think), the point of $R$ is just for the external view in triangle $\triangle AIB$ since after this we can construct the angle bisectors of $\angle BAI$ and $\angle BIR$, which leads to excenter of $AIB$. Edited! Cool, you got the same solution

HoRI_DA_GRe8 wrote: So nice Ok so here is my solution from contest First draw the incentre $I$ of $\triangle BIC$.Mark the lines $BI$ and $CI$.Let $\ell_{XYZ}$ denote the angle bisector of $\angle XYZ$. First draw $\ell_{BAI}$ and mark $CI \cap \ell_{BAI}=E$ .Clearly $E$ lies on the opposite side of $I$ w.r.t $C$ Now draw $\ell_{EIB}$ and mark $\ell_{EIB}\cap \ell_{ICB}=K_C$.Note that $K_C$ is the $C$-excentre of $\triangle BIC$.Similarly construct $K_B$ which is the $B$ excentrez of $\triangle BIC$. Now note that $K_BCBK_C$ is cyclic.So by fact $5$ we have that $\ell_{K_BBK_C} \cap \ell_{K_BCK_C}=M$ lie on the circle $(K_BCBK_C)$ and $M$ is also the midpoint of arc $K_BK_C$ (on which $B$,$C$ doesn't lie). Now since $\triangle MK_BK_C$ is isosceles, we have that $\ell_{K_BMK_C}$ is perpendicular bisector of $K_BK_C$ , since $K_BK_C$ is nothing other than $\ell_{EIB}$ , we can mark $\ell_{K_BMK_C}\cap K_BK_C=L$ .Now note that $L$ is the centre of $(K_BCBK_C)$ Thus we have $\ell_{BLC}$ is the perpendicular bisector of $BC$ This we have drawn the perpendicular bisector of $BC$.Repeat this process with $CA$ or $AB$ and mark the intersection of the perpendicular bisectors to get the circumcentre $\blacksquare$ Amazing solution bro, loved it

Seems like no one has posted my solution yet, which I'd like to think is one of the simplest Obviously we can construct $I$, the incenter of $\triangle ABC$. We now make the key claim Claim: In $\triangle ABC$, we can construct the midpoint of minor arc $BC$ given a point $D$ lying on ray $AB$ on the opposite side of $A$ Proof: Let $I$ denote the incenter of $\triangle ABC$. Construct the $A$-excenter, $I_A$ of $\triangle ABC$ by marking the intersections of $\angle DBC$ and $\angle BAC$. By the incenter excenter lemma of course $IBI_AC$ is cyclic. We can construct the mipdoint of arc of minor arc $CI_A$ in ($IBI_AC$) (say $M_C$) by marking intersections of $I_ABC$ and $I_AIC$. Now we just take intersection of angle bisector of $\angle CM_CI_A$ and $AI_A$ to get the midpoint of minor arc $BC$ in $(ABC)$ $\square$ Back to the main problem Let $D$ be the intersection of angle bisector of $\angle ABI$ and $CI$. This is clearly an external point on ray $CI$, so we use our claim to construct the midpoint of arc $AI$ in ($AIC$), say $X$. Notice that the intersection of angle bisector of $\angle AXI$ and $CI$ is the arc midpoint of minor arc $AB$ in $(ABC)$ Now we construct the perpendicular bisector of $AB$ and repeat the same process to get the perpendicular bisector of $AC$, which ultimately gives us the circumcenter of $(ABC)$ $\blacksquare$ Thanks a LOT Shankadeep for proposing a geometry, without this we would have a 0 geo INMO with 5 combinatorics problems

BVKRB- wrote: Seems like no one has posted my solution yet, which I'd like to think is one of the simplest Obviously we can construct $I$, the incenter of $\triangle ABC$. We now make the key claim Claim: In $\triangle ABC$, we can construct the midpoint of minor arc $BC$ given a point $D$ lying on ray $AB$ on the opposite side of $A$ Proof: Let $I$ denote the incenter of $\triangle ABC$. Construct the $A$-excenter, $I_A$ of $\triangle ABC$ by marking the intersections of $\angle DBC$ and $\angle BAC$. By the incenter excenter lemma of course $IBI_AC$ is cyclic. We can construct the mipdoint of arc of minor arc $CI_A$ in ($IBI_AC$) (say $M_C$) by marking intersections of $I_ABC$ and $I_AIC$. Now we just take intersection of angle bisector of $\angle CM_CI_A$ and $AI_A$ to get the midpoint of minor arc $BC$ in $(ABC)$ $\square$ Back to the main problem Let $D$ be the intersection of angle bisector of $\angle ABI$ and $CI$. This is clearly an external point on ray $CI$, so we use our claim to construct the midpoint of arc $AI$ in ($AIC$), say $X$. Notice that the intersection of angle bisector of $\angle AXI$ and $CI$ is the arc midpoint of minor arc $AB$ in $(ABC)$ Now we construct the perpendicular bisector of $AB$ and repeat the same process to get the perpendicular bisector of $AC$, which ultimately gives us the circumcenter of $(ABC)$ $\blacksquare$ Thanks a LOT Shankadeep for proposing a geometry, without this we would have a 0 geo INMO with 5 combinatorics problems Amazing solution

Fun fact- This tool can do the work of a straightedge too.

Solve using geometry

Really elegant and interesting