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Quad formed by orthocenters has same area (all 7's!)
v_Enhance   35
N 9 minutes ago by Wictro
Source: USA January TST for the 55th IMO 2014
Let $ABCD$ be a cyclic quadrilateral, and let $E$, $F$, $G$, and $H$ be the midpoints of $AB$, $BC$, $CD$, and $DA$ respectively. Let $W$, $X$, $Y$ and $Z$ be the orthocenters of triangles $AHE$, $BEF$, $CFG$ and $DGH$, respectively. Prove that the quadrilaterals $ABCD$ and $WXYZ$ have the same area.
35 replies
v_Enhance
Apr 28, 2014
Wictro
9 minutes ago
A Segment Bisection Problem
buratinogigle   2
N 19 minutes ago by aidenkim119
Source: VN Math Olympiad For High School Students P9 - 2025
In triangle $ABC$, let the incircle $\omega$ touch sides $BC, CA, AB$ at $D, E, F$, respectively. Let $P$ lie on the line through $D$ perpendicular to $BC$. Let $Q, R$ be the intersections of $PC, PB$ with $EF$, respectively. Let $K, L$ be the projections of $R, Q$ onto line $BC$. Let $M, N$ be the second intersections of $DQ, DR$ with the incircle $\omega$. Let $S$ be the intersection of $KM$ and $LN$. Prove that the line $DS$ bisects segment $QR$.
2 replies
buratinogigle
Today at 1:36 AM
aidenkim119
19 minutes ago
No more topics!
Probably appeared before
steven_zhang123   3
N Mar 30, 2025 by steven_zhang123
In the plane, there are two line segments $AB$ and $CD$, with $AB \neq CD$. Prove that there exists and only exists one point $P$ such that $\triangle PAB \sim \triangle PCD$.($P$ corresponds to $P$, $A$ corresponds to $C$)
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3 replies
steven_zhang123
Mar 30, 2025
steven_zhang123
Mar 30, 2025
Probably appeared before
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steven_zhang123
411 posts
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In the plane, there are two line segments $AB$ and $CD$, with $AB \neq CD$. Prove that there exists and only exists one point $P$ such that $\triangle PAB \sim \triangle PCD$.($P$ corresponds to $P$, $A$ corresponds to $C$)
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This post has been edited 2 times. Last edited by steven_zhang123, Mar 30, 2025, 4:21 AM
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lyllyl
4 posts
#2
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I think the problem is wrong.For example,assumn A:(-1,2) B:(-1,0) C:(2,2) D:(4,1) So P can be (1,0) and (0.2, 4.4),not only one!
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whwlqkd
97 posts
#3
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Isn’t it Miquel Point?
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steven_zhang123
411 posts
#4
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@above, yes. But do you have a solution that doesn't require the use of Miquel point and circles?
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