Circumcenter of a Tetrahedron

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Circumcenter of a Tetrahedron

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Source: All-Russian MO 2001 Grade 11 #8

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#1 h Jan 3, 2012, 12:39 AM • 5 Y

Y by HamstPan38825, Adventure10, HHGB, MS_asdfgzxcvb, and 1 other user

A sphere with center on the plane of the face $ABC$ of a tetrahedron $SABC$ passes through $A$ , $B$ and $C$ , and meets the edges $SA$ , $SB$ , $SC$ again at $A_1$ , $B_1$ , $C_1$ , respectively. The planes through $A_1$ , $B_1$ , $C_1$ tangent to the sphere meet at $O$ . Prove that $O$ is the circumcenter of the tetrahedron $SA_1B_1C_1$ .

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#2 h Jan 3, 2012, 5:26 AM • 4 Y

Y by Adventure10, Mango247, HHGB, MS_asdfgzxcvb

Let $\mathcal{S}$ denote the sphere passing through $A,B,C$ and centered on the plane $ABC.$ Inversion with center $S$ and power $SA \cdot SA_1=SB \cdot SB_1=SC \cdot SC_1$ takes $\mathcal{S}$ into itself and swaps the circumsphere $\mathcal{O}$ of $SA_1B_1C_1$ and the plane $ABC.$ The angle between $\mathcal{S}$ and the plane $ABC$ is right, thus by conformity the angle between $\mathcal{S}$ and $\mathcal{O}$ is also right, i.e. $\mathcal{S}$ and $\mathcal{O}$ are orthogonal $\Longrightarrow$ Tangent planes of $\mathcal{S}$ at $A_1,B_1,C_1$ pass through the center of $\mathcal{O}.$

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#3 h Mar 31, 2017, 6:21 PM • 3 Y

Y by Adventure10, MS_asdfgzxcvb, HHGB

Let $\mathcal{P}$ denote the plane of triangle $ABC$ , $X$ be the circumcenter of $\triangle ABC$ (also the center of the sphere $\mathcal{S}$ passing through $A, B, C, A_1, B_1, C_1$ ) and redefine $O$ as the center of the circumsphere of tetrahedron $SA_1B_1C_1$ . I will show that $O$ lies on the plane $\mathcal{P}_A$ passing through $A_1$ and tangent to $\mathcal{S}$ . Along with cyclic variants for $B_1, C_1$ this will prove our result.

Let $S'$ be the point symmetric to $S$ about the plane $\mathcal{P}$ . Apply inversion around a sphere centered at $S$ having radius $\sqrt{SA \cdot SA_1}$ . Note that $\mathcal{S}$ is fixed by the map and $\mathcal{P}$ is mapped to the circumsphere of $SA_1B_1C_1$ . Hence, $O$ is mapped to $S'$ and $\mathcal{P}_A$ is mapped to a sphere passing through $A, S$ and tangent to $\mathcal{S}$ . Let $O_A$ be the center of this sphere, and observe that $A, X, O_A$ are collinear by dilation at $A$ . Evidently, $O_A$ lies in the plane $\mathcal{P}$ so $O_AS=O_AS'$ and so $S'$ lies on the image of $\mathcal{P}_A$ under the inversion. Our claim holds and the result follows. $\square$

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#4 h May 27, 2019, 8:17 PM • 3 Y

Y by Adventure10, Mango247, HHGB

Here is a dumber solution than the above

A bit too easy for a Russian 3-D Geo, especially as a #8

Let $P$ be the foot from $S$ to the plane of $ABC.$ Let $O_1$ be the center of the sphere. Consider the inversion $\gamma$ at $S$ with power $SA \cdot SA_1.$ Observe that $A \rightarrow A_1, B \rightarrow B_1, C \rightarrow C_1$ by Power of the Point. Now, let $\gamma (P) = X,$ and $O$ be the midpoint of $SX.$ We claim that $O$ is the circumcenter of tetrahedron $SA_1B_1C_1$ and also satisfies $\angle OA_1O_1 = \angle OB_1O_1 = \angle OC_1O_1,$ from which the problem would follow. Observe that $SA_1 \cdot SA = SX \cdot SP$ by definition, and so since $S, A_1, A, P, X$ are all coplanar, we have that $\triangle SA_1X \sim \triangle SPA.$ Hence, as $\angle SPA = 90$ we know that $\angle SA_1 X = 90.$ Hence, as $O$ is the midpoint of the hypotenuse of right $\triangle SA_1X$ , we have that $OA_1 = OS.$ Similarly, $OB_1 = OS, OC_1 = OS$ , and so we've shown that $O$ is the circumcenter of tetrahedron $SA_1B_1C_1.$

Hence, it only suffices to verify that $\angle OA_1 O_1 = 90$ , as the other two right angles would follow by symmetry. We need to show that $OA_1 ^2 + O_1A_1^2 = OO_1^2.$ Observe that
$OO_1^2 - OA_1^2 = OO_1^2 - OS^2 = O_1P^2 + PO^2 - OS^2 = O_1P^2 + (PO-OS)(PO+OS),$ which implies that $OO_1^2 - OA_1^2 = O_1P^2 + (PO-OX)(SP) = O_1P^2 + SP \cdot PX.$
From this, we have that $OO_1^2 - OA_1^2 = O_1P^2 + SP^2 - SX \cdot SP = O_1S^2 - SA_1 \cdot SA = O_1A^2,$ where the last part follows from Power of the Point at $S.$

$\square$

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HHGB

#5 h Jan 25, 2025, 12:41 PM

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Let the sphere mentioned in the problem be $\Gamma$ .
Let $H$ be the foot of S to the plane of $ASH$ .
Let $\omega$ be the the intersection of the plane of $ASH$ and the sphere.
The tangent plane through $A_1$ is also tangent to $\omega$ . Let $X$ be the intersection of the tangent plane of $A_1$ and $SH$ . Looking from the perspective of the plane of $ASH$ and by angle chasing, $SX=A_{1}X$ .
Let $\angle SAH=\alpha$ .
$\frac{SA_1}{2SX}=\sin \alpha=\frac{SH}{SA}$
$\Rightarrow SX=\frac{SA \cdot SA_1}{2SH}=\frac{{\Pi}_{\omega}(S)}{2SH}=\frac{{\Pi}_{\Gamma}(S)}{2SH}$
$=\frac{SB \cdot SB_1}{2SH}=\frac{SC \cdot SC_1}{2SH}$ .
Redefining $X$ for $B$ and $C$ implies that all three planes intersect $SH$ at a point whose distance from $S$ is $\frac{{\Pi}_{\Gamma}(S)}{2SH}$ which is symmetric about $A$ , $B$ and $C$ .
Thus, $O$ lies on $SH$ , and:
$OA_1=OB_1=OC_1=OS$ .
So $O$ is the circumcenter of the tetrahedron $SABC$ .
$\Box$

This post has been edited 3 times. Last edited by HHGB, Feb 11, 2025, 8:17 PM
Reason: Fixing some misnaming

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