Ratio of lengths in right-angled triangle

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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

#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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#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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#3 h Apr 8, 2025, 4:26 PM • 1 Y

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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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