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Gheorghe Țițeica 2025 Grade 8 P3
AndreiVila   1
N 19 minutes ago by sunken rock
Source: Gheorghe Țițeica 2025
Two regular pentagons $ABCDE$ and $AEKPL$ are given in space, such that $\angle DAC = 60^{\circ}$. Let $M$, $N$ and $S$ be the midpoints of $AE$, $CD$ and $EK$. Prove that:
[list=a]
[*] $\triangle NMS$ is a right triangle;
[*] planes $(ACK)$ and $(BAL)$ are perpendicular.
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Ukraine Olympiad
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AndreiVila
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sunken rock
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equal angles
jhz   3
N Mar 26, 2025 by DottedCaculator
Source: 2025 CTST P16
In convex quadrilateral $ABCD, AB \perp AD, AD = DC$. Let $ E$ be a point on side $BC$, and $F$ be a point on the extension of $DE$ such that $\angle ABF = \angle DEC>90^{\circ}$. Let $O$ be the circumcenter of $\triangle CDE$, and $P$ be a point on the side extension of $FO$ satisfying $FB =FP$. Line BP intersects AC at point Q. Prove that $\angle AQB =\angle DPF.$
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jhz
Mar 26, 2025
DottedCaculator
Mar 26, 2025
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equal angles
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Source: 2025 CTST P16
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jhz
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jhz
#1 h  Mar 26, 2025, 12:56 AM
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In convex quadrilateral $ABCD, AB \perp AD, AD = DC$. Let $ E$ be a point on side $BC$, and $F$ be a point on the extension of $DE$ such that $\angle ABF = \angle DEC>90^{\circ}$. Let $O$ be the circumcenter of $\triangle CDE$, and $P$ be a point on the side extension of $FO$ satisfying $FB =FP$. Line BP intersects AC at point Q. Prove that $\angle AQB =\angle DPF.$
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jhz
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jhz
#2 h  Mar 26, 2025, 2:58 AM
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Reconstruct $P$ as the intersection of $\odot ABC$ and $\odot DEC$. remain is simple calculation.
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YaoAOPS
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YaoAOPS
#3 h  Mar 26, 2025, 3:39 AM
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Sketch since its so painful.

Redefine $P$ as $(ABC) \cap (DEC)$. Then we may show that $\angle FBP = \angle BPO$ through angle chasing, redefine $F$ as $BF \cap PO$. Then let $H = BA \cap ED$, note that $(BEH)$ is cyclic and tangent to $BF'$, likewise we may angle chase that $DAHP$ is cyclic so $\angle DPH = 90^\circ$ and thus $(EPH)$ is tangent to $F'P$, thus $F' \in EDH$, and thus $F' = F$. Finally, the result is another painful angle chase.
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DottedCaculator
7319 posts
DottedCaculator
#4 h  Mar 26, 2025, 6:51 PM
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this is why i'm bad at geo

Claim: $P$ lies on $(CDE)$

Proof: Verify $\sqrt{FE\cdot FD+R^2}=BF-R$, where $R$ is the radius of $(CDE)$ by trig bash

Now, construct $P'=FO\cap(CDE)$ and then angle/arc bash the required angle equality.
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