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4 lines concurrent
Zavyk09   3
N 13 minutes ago by ItzsleepyXD
Source: Homework
Let $ABC$ be triangle with circumcenter $(O)$ and orthocenter $H$. $BH, CH$ intersect $(O)$ again at $K, L$ respectively. Lines through $H$ parallel to $AB, AC$ intersects $AC, AB$ at $E, F$ respectively. Point $D$ such that $HKDL$ is a parallelogram. Prove that lines $KE, LF$ and $AD$ are concurrent at a point on $OH$.
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Zavyk09
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ItzsleepyXD
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Ratio of lengths in right-angled triangle
DylanN   2
N Tuesday at 4:26 PM by sunken rock
Source: South African Mathematics Olympiad 2021, Problem 2
Let $PAB$ and $PBC$ be two similar right-angled triangles (in the same plane) with $\angle PAB = \angle PBC = 90^\circ$ such that $A$ and $C$ lie on opposite sides of the line $PB$. If $PC = AC$, calculate the ratio $\frac{PA}{AB}$.
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DylanN
Aug 11, 2021
sunken rock
Tuesday at 4:26 PM
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Ratio of lengths in right-angled triangle
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Source: South African Mathematics Olympiad 2021, Problem 2
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DylanN
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DylanN
#1 h  Aug 11, 2021, 2:59 PM
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Let $PAB$ and $PBC$ be two similar right-angled triangles (in the same plane) with $\angle PAB = \angle PBC = 90^\circ$ such that $A$ and $C$ lie on opposite sides of the line $PB$. If $PC = AC$, calculate the ratio $\frac{PA}{AB}$.
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Mathzeus1024
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Mathzeus1024
#2 h  Apr 7, 2025, 12:24 PM
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Let $a,b>0$ such that $P(a,b); A(0,b); B(0,0)$ exist in the first quadrant of the $xy-$plane for right $\Delta PAB$. If hypothenuse $BP$ is represented by the line $y=\frac{b}{a}x$, then let side $BC$ of similar right $\Delta PBC$ be represented by $y = -\frac{a}{b}x$ such that $C\left(t, -\frac{at}{b}\right)$. We are given that $|PC|=|AC|$, which satisfies:

$\sqrt{(a-t)^2+\left(b+\frac{at}{b}\right)^2} = \sqrt{t^2 + \left(b+\frac{at}{b}\right)^2} \Rightarrow t =\frac{a}{2}$

or $C\left(\frac{a}{2}, -\frac{a^2}{2b}\right)$. Next, let:

$|PB|=k|PA| \Rightarrow \sqrt{a^2+b^2}= ka$ (i);

$|PC|=k|PB| \Rightarrow \sqrt{\left(a-\frac{a}{2}\right)^2+\left(b+\frac{a^2}{2b}\right)^2} = k\sqrt{a^2+b^2}$ (ii)

of which we obtain:

$\frac{a^2+b^2}{a} = \sqrt{\left(a-\frac{a}{2}\right)^2+\left(b+\frac{a^2}{2b}\right)^2}$;

or $\frac{(a^2+b^2)^2}{a^2} = \left(\frac{a}{2}\right)^2 + \left(\frac{2b^2+a^2}{2b}\right)^2$;

or $4a^4b^2+8a^2b^4+4b^6 = a^4b^2+4a^2b^4+4a^4b^2+a^6$;

or $0 = (a^2+b^2)(a^2+2b^2)(a+\sqrt{2}b)(a-\sqrt{2}b) \Rightarrow a=\sqrt{2}b$ (iii).

Hence, $\frac{|PA|}{|AB|} = \frac{a}{b} = \textcolor{red}{\sqrt{2}}$.
This post has been edited 7 times. Last edited by Mathzeus1024, Apr 7, 2025, 12:38 PM
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sunken rock
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sunken rock
#3 h  Tuesday at 4:26 PM
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My solution with less calculations at https://stanfulger.blogspot.com/2025/04/aops-httpsartofproblemsolvingcomcommuni.html.

Best regards,
sunken rock
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