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#1 h Apr 19, 2025, 7:37 AM

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Classify all possible quadrupes of $4$ distinct points in a plane such the circumradius of any $3$ of them is the same.

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#2 h Apr 19, 2025, 10:40 AM

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Fakesolve

This post has been edited 1 time. Last edited by CHESSR1DER, Apr 20, 2025, 7:22 PM

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#4 h Apr 19, 2025, 1:27 PM

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ThatApollo777 wrote:

Classify all possible quadruples of $4$ distinct points in a plane such the circumradius of any triangle formed by three of them is the same.

Let $\triangle ABC$ be an acute triangle, and let $H$ be an arbitrary point that satisfies the given conditions. (The same reasoning can be adapted for obtuse or right triangles, with the only exception that H lies on or outside the triangle.)

Let $\angle A, \angle B, \angle C$ denote the interior angles of $\triangle ABC$ . Let $O$ be the circumcenter of $\triangle ABC$ , and let $(O)$ denote the circumcircle centered at $O$ that passes through $A, B, C$ .

If $H$ lies outside of $(O)$ , then $\angle BHC < \angle A$ , so the radius of the circle through $BHC$ (denote this circle as $(O_A)$ ) would be different from the radius of $(O)$ , which is a contradiction since we're assuming congruent cyclic configurations. Hence, $H$ must lie on or inside $(O)$ .

Obviously, if $A, B, C, H$ are concyclic, then the claim is satisfied.

Now, consider the case where $H$ lies strictly inside $(O)$ . Suppose $H$ lies outside or on the boundary of triangle $ABC$ . Without loss of generality, assume $H$ and $A$ lie on opposite sides of segment $BC$ . Then, since $A$ and $H$ lie on different sides of $BC$ , we must have:
$\angle BAC + \angle BHC = 180^\circ.$ But this is impossible, since $H$ lies within the circle $(O)$ , so $\angle BAC + \angle BHC > 180^\circ$ . This contradiction shows that $H$ must lie inside $\triangle ABC$ .

From this, it follows:
$\angle BHC = 180^\circ - \angle A, \quad \angle CHA = 180^\circ - \angle B, \quad \angle AHB = 180^\circ - \angle C.$
Let the lines $AH, BH, CH$ intersect $BC, CA, AB$ at points $D, E, F$ respectively. Then, the quadrilaterals $AEHF, BFHD, CDHE$ are all cyclic.

Now consider $\angle DHE$ . We have:
$\angle DHE = \angle CHD + \angle CHE = (180^\circ - \angle AHC) + (180^\circ - \angle BHC) = \angle A + \angle B.$ This implies $\angle AFE = \angle AHE = \angle C$ . Therefore, quadrilateral $BCEF$ is cyclic, implying $\angle BEC = \angle BFC$ , and likewise, $\angle AEH = \angle AFH$ . Since:
$\angle AEH + \angle AFH = 180^\circ,$ each of these angles must be $90^\circ$ , implying that $BH$ and $CH$ are altitudes of triangle $ABC$ . Hence, $H$ is the orthocenter of $\triangle ABC$ .

In total, there are two possible configurations of the four points:
1. Four points that lie on a circle (i.e., they form a concyclic quadrilateral).
2. Four points such that one of them is the orthocenter of the triangle formed by the other three.

This post has been edited 3 times. Last edited by pooh123, Apr 19, 2025, 1:33 PM
Reason: typo

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#5 h Apr 20, 2025, 11:28 AM

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CHESSR1DER wrote:

Let $ABC$ is a triangle from first $3$ points. Let $O$ be circumcenter of $ABC$ .
Let $O_a,O_b,O_c$ be circumcenters of $BCD,ACD,ABD$ . Then it's easy to see that $O_a,O_b,O_c$ are reflections of $O$ through $BC,AC,AB$ respectively. So circles are constant. Then $D$ is the intersection of $(ABD),(ACD),(BCD)$ . Easy to see that $O_a,O_b,O_c$ are different points, so circles dont coincide. So we need to check when $(ABD),(ACD),(BCD)$ meet in $1$ point. Since $3$ equal circles meet in $1$ point $(ABC)$ is equilateral. So $D=O$ .
Answer: $ABC$ forms equilateral triangle and $D$ is circumcenter of $ABC$ .

This is not correct unfortunately

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