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In $\Delta ABC$, $\angle ABC = 90^\circ$ , $D$ is the midpoint of line $BC$. Point $P$ is on $AD$ line. $PM$ & $PN$ are respectively perpendicular on $AB$ & $AC$. $PM$ = $2PN$, $AB = 5$, $BC$ = $a\sqrt{b}$, where $a, b$ are positive integers. $a+b$ = ?

Source: Bangladesh Math Olympiad 2018 junior category.

I could not find any way to relate $PM$ = $2PN$ to this math.

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  • $\begingroup$ This may not be unique. What if $|BC|^2$ were $75$, do we take $a=1$ or $a=5$? Or is the correct value amenable to $a=1$ only? $\endgroup$ – Oscar Lanzi Jan 17 at 14:18
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Diagram

$\triangle ABC \sim \triangle AMM' \sim \triangle PNM'$

$\frac{BC}{AB}=\frac{NM'}{PN}$

$PM'=PM=2PN$

$NM'=\sqrt{(2PN)^2-PN^2}=PN\sqrt3$

$\frac{NM'}{PN}=\sqrt3$

$\frac{BC}{AB}=\sqrt3$

$BC=AB\sqrt3=5\sqrt3$

$a=5,b=3,\color{blue}{a+b=8}$

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  • $\begingroup$ That's strange, I thought the sum would be 76. $\endgroup$ – Oscar Lanzi Jan 17 at 15:02
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    $\begingroup$ @OscarLanzi As you pointed out, $5\sqrt3=1\sqrt{75}$, so $(a,b)=(1,75)$ technically qualifies. $\endgroup$ – Daniel Mathias Jan 17 at 15:06
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Let $$\angle{BAC}=\alpha$$ then $$\tan(\alpha)=\frac{PM}{5-PN}=-\frac{PN}{\frac{a}{2}-2PN}$$ so we get $$\frac{a}{2}+5=3PN$$ (1) and $$AD^2=5^2+\frac{a^2}{4}$$ so $$\left(\frac{2PN}{\sin(\alpha)}+\frac{PN}{\cos(\alpha)}\right)^2=5^2+\frac{a^2}{4}$$ and for $PN$ we have $$PN=\frac{1}{3}\left(\frac{a}{2}+5\right)$$ and $$\cos^2(\alpha)=\frac{1}{\tan^2(\alpha)+1)}$$ and $$\sin^2(\alpha)=\frac{\tan^2(\alpha)}{1+\tan^2(\alpha)}$$ Can you finish?

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We can think that $B=(0,0)$. Then $A=(0,5)$ and $C=(a\sqrt{b},0)$.

Then we have that $D=(\frac{a\sqrt{b}}{2},0)$.

Since $P$ is in $AD$, $P=(0,5)+\lambda_1 (a\sqrt{b},-10)$.

Now, as we can see in the picture, we do not know where $N$ and $M$ are. But by what you said, we have two options for each one:

$N_1= (0,5)+\lambda_2(a\sqrt{b},-5)$

$N_2= (0,0)+\lambda_3(a\sqrt{b},0) = (\lambda_3 \cdot a\sqrt{b},0)$

$M_1= (0,0)+\lambda_4(0,5)=(0,\lambda_4 \cdot 5)$

$M_2= (0,5)+\lambda_5(a\sqrt{b},-5)$

Now, we have to play computating $PM$ and $PN$ and using your relation.

enter image description here

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  • $\begingroup$ I will edit it now. Thanks! $\endgroup$ – idriskameni Jan 17 at 13:55
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Draw the triangle as instructed. $M$ and $N$ are taken to be the feet if the respective perpendiculars to the triangle's sides, $M$ on $AB$ and $N$ on $AC$.

Draw $DN'$ parallel to $PN$ with $N'$ also on $AC$. Then $\triangle AMP ~ \triangle ABD$ and $\triangle APB ~ \triangle ADN'$, forcing $|PM|/|PN|=|BD|/|DN'|=|CD|/|DN'|=2$. Then right $\triangle DN'C$ has a hypoteneyse/leg ratio of $2:1$ hence $\angle C$ measures $30°$.

It follows that right $\triangle ABC$ with a $30°$ angle at $C$ also has a $2:1$ ratio of the hypotenuse to the shorter leg, therefore its hypotenuse is $10$ and $|BC|=5\sqrt{3}=1\sqrt{75}$. And the solution is nonunique! C'mon man!

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