# What is the value of $a + b$ if $a\sqrt{b} = BC$ in right triangle $\Delta ABC$?

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. $$a+b$$ = ?

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

• 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? – Oscar Lanzi Jan 17 '19 at 14:18

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

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

$$PM'=PM=2PN$$

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

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

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

$$BC=AB\sqrt3=5\sqrt3$$

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

• That's strange, I thought the sum would be 76. – Oscar Lanzi Jan 17 '19 at 15:02
• @OscarLanzi As you pointed out, $5\sqrt3=1\sqrt{75}$, so $(a,b)=(1,75)$ technically qualifies. – Daniel Mathias Jan 17 '19 at 15:06

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?

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.

• I will edit it now. Thanks! – idriskameni Jan 17 '19 at 13:55

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!