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Line passes through fixed point, as point varies
Jalil_Huseynov   60
N 21 minutes ago by Rayvhs
Source: APMO 2022 P2
Let $ABC$ be a right triangle with $\angle B=90^{\circ}$. Point $D$ lies on the line $CB$ such that $B$ is between $D$ and $C$. Let $E$ be the midpoint of $AD$ and let $F$ be the seconf intersection point of the circumcircle of $\triangle ACD$ and the circumcircle of $\triangle BDE$. Prove that as $D$ varies, the line $EF$ passes through a fixed point.
60 replies
Jalil_Huseynov
May 17, 2022
Rayvhs
21 minutes ago
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Tangent to two circles
Mamadi   2
N 35 minutes ago by A22-
Source: Own
Two circles \( w_1 \) and \( w_2 \) intersect each other at \( M \) and \( N \). The common tangent to two circles nearer to \( M \) touch \( w_1 \) and \( w_2 \) at \( A \) and \( B \) respectively. Let \( C \) and \( D \) be the reflection of \( A \) and \( B \) respectively with respect to \( M \). The circumcircle of the triangle \( DCM \) intersect circles \( w_1 \) and \( w_2 \) respectively at points \( E \) and \( F \) (both distinct from \( M \)). Show that the line \( EF \) is the second tangent to \( w_1 \) and \( w_2 \).
2 replies
Mamadi
May 2, 2025
A22-
35 minutes ago
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Geometry
Lukariman   7
N 2 hours ago by vanstraelen
Given circle (O) and point P outside (O). From P draw tangents PA and PB to (O) with contact points A, B. On the opposite ray of ray BP, take point M. The circle circumscribing triangle APM intersects (O) at the second point D. Let H be the projection of B on AM. Prove that $\angle HDM$ = 2∠AMP.
7 replies
Lukariman
Yesterday at 12:43 PM
vanstraelen
2 hours ago
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two 3D problems in one day
egxa   1
N Apr 18, 2025 by sami1618
Source: All Russian 2025 11.2
A right prism \(ABCA_1B_1C_1\) is given. It is known that triangles \(A_1BC\), \(AB_1C\), \(ABC_1\), and \(ABC\) are all acute. Prove that the orthocenters of these triangles, together with the centroid of triangle \(ABC\), lie on the same sphere.
1 reply
egxa
Apr 18, 2025
sami1618
Apr 18, 2025
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two 3D problems in one day
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Source: All Russian 2025 11.2
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egxa
210 posts
egxa
#1 h  Apr 18, 2025, 9:44 AM • 2 Y
Y by buratinogigle, Miquel-point
A right prism \(ABCA_1B_1C_1\) is given. It is known that triangles \(A_1BC\), \(AB_1C\), \(ABC_1\), and \(ABC\) are all acute. Prove that the orthocenters of these triangles, together with the centroid of triangle \(ABC\), lie on the same sphere.
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sami1618
907 posts
sami1618
#2 h  Apr 18, 2025, 2:41 PM
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Let $H_A$, $H_B$, $H_C$, and $H$ be the orthocenters of triangles $A_1BC$, $AB_1C$, $ABC_1$, and $ABC$, respectively. Let $G$ be the centroid of triangle $ABC$.

Claim. The point $H_A$ is the foot of $H$ onto the plane $A_1BC$.
Proof. Let $H'$ be the foot of $H$ onto the plane $A_1BC$. Notice that $BH$ is perpendicular to plane $ACC_1A_1$. Thus $BH\perp A_1C$, which implies $BH'\perp A_1C$. Similarly, $CH'\perp A_1B$. Thus $H'=H_A$, as required.

An identical result holds for $H_B$ and $H_C$. Notice the centroids of triangles $A_1BC$, $AB_1C$, and $ABC_1$ all coincide at some point $X$. Since $G$ is the foot of $X$ onto the plane $ABC$, it follows that $\angle HGX=90^{\circ}$. By the claim, we also get that $\angle HH_AX=HH_BX=HH_CX=90^{\circ}$. Thus all five points lie on the sphere with diameter $HX$, concluding the proof.
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