What's tha measure of the segment $AC$ in the question below?

Question

For reference:In triangle ABC ($<B = 90°$) the median $BM$ is drawn. Through the midpoint $N$ of this median a straight line perpendicular to $AC$ is drawn which intercepts $AC$ and $BC$ at points $Q$ and $P$. If: $PQ=5, QN=4$, calculate $AC$ (answer: $24$)

My progress..

$MC = MB = MA = R\\ \triangle BMC (isosceles) \implies BD = BC$

I try T. Menelaus Th.

$\triangle PQC : MB\\ CM\cdot QN \cdot PB = MQ \cdot PN \cdot. CB\\ R.\cdot 4 \cdot PB = MQ \cdot 1 \cdot CB \implies\\$

I think the solution is here...

Answer 1

$MN=x$, $QC=y$, $QM=y-2x$, $(y-2x)^2+QN^2=x^2$

$CP=\sqrt{QP^2+y^2}$

$S_{QCP}=\frac{QP}{2}y$

$h_{ABC}=2QN\Rightarrow S_{ABC}=4x\cdot QN$

$\frac{S_{ABC}}{S_{QCP}}=\frac{AC^2}{CP^2}=\frac{16x^2}{QP^2+y^2}=\frac{8x\cdot QN}{y\cdot QP}\Rightarrow x=\frac{(QP^2+y^2)\cdot QN}{2y\cdot QP}$

$(y-2x)^2+QN^2=x^2\Rightarrow\left(y-\frac{(QP^2+y^2)\cdot QN}{y\cdot QP}\right)^2+QN^2=\left(\frac{(QP^2+y^2)\cdot QN}{2y\cdot QP}\right)^2\Rightarrow$ $(2y^2\cdot QP-2(QP^2+y^2)\cdot QN)^2+4y^2\cdot QN^2\cdot QP^2=(QP^2+y^2)^2\cdot QN^2$

This is biquadratic equation that can be solved analytically but general formula for solution is too long.

If $QN=4$, $QP=5$: $(10y^2-200-8y^2)^2+1600y^2=16(25+y^2)^2\Rightarrow$ $y^4-200y^2+10000+400y^2=4y^4+200y^2+2500\Rightarrow$ $3y^4-7500=0 \Rightarrow y=5\sqrt{2} \Rightarrow x=\frac{(25+50)\cdot 4}{2\cdot 5\sqrt{2}\cdot 5}=3\sqrt{2}$.

But this result doesn't have sense, because $y$ must be greater than $2x$.

When taking $NP=1.55$, $QN=2.9$ the answer is $AC=4x=23.6$

Answer 2

Since the given data are inconsistent, I will present a general solution.

Consider the image below.

Say $\small EF=m$ and $\small EH=n$. Prolong $FH$ such that it meets the line through $\small B$ parallel to $\small AC$ at $\small G$. We can see that $\small \triangle EFD$ and $\small \triangle EGB$ are congruent. Therefore $\small BG=FD=x$ and $\small GH=(m-n)$.

$$\small \triangle CHF\sim\triangle BHG\implies\frac {R+x}x=\frac{m+n}{m-n}$$ $$\small \therefore x=\frac{(m-n)R}{2n}$$

As $\small ED=R/2$, we can apply Pythagorean theorem to $\small \triangle EFD$ which gives, $$\small m^2+\left(\frac{(m-n)R}{2n}\right)^2=\left(\frac R2\right)^2$$ This simplifies to $$\small (m^2-2mn)R^2+4m^2n^2=0$$ In the case of $\small m=4$ and $\small n=1$, we get $\small R=\sqrt{-8}$ which is undefined.

According to WolframAlpha, we can find these positive integer solutions those satisfy the above equation.