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in $\Delta ABC$,if $AD$ bisects $\angle BAC$, $MD$ bisects $\angle ADB$, $ND$ bisects $\angle ADC$,prove that $$\dfrac{1}{BM}+\dfrac{1}{CN}\le\dfrac{4}{MN}$$

Illustration

my idea:use if $AD$ bisects $\angle BAC$,then we have $$\dfrac{AB}{AC}=\dfrac{BD}{DC}$$

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  • $\begingroup$ yes, I have edit $\endgroup$
    – math110
    Aug 6, 2013 at 12:09

3 Answers 3

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Your statement cannot be proven because it is incorrect. Take the following coordinates:

\begin{align*} A &= \begin{pmatrix}0\\1\end{pmatrix} & B &= \begin{pmatrix}1\\0\end{pmatrix} & C &= \begin{pmatrix}\lambda\\0\end{pmatrix} \end{align*}

Then in the limit $\lambda\to+\infty$ you get

$$\lim_{\lambda\to+\infty} \frac{MN}{BM} + \frac{MN}{CN} = \lim_{\lambda\to+\infty}\frac{MN}{BM} = \sqrt{\sqrt{4\sqrt2 + 8} + 4\sqrt2 + 8} \approx 4.1656 > 4 $$

This result can either be obtained in a CAS of your choice, or – if you don't need the exact number and trust your floating point computations – in some suitable geometry program. In either case, the bottom line is that if you just move $C$ far enough, and have $\angle CBA$ sufficiently large, then the equation you stated will be violated.

Illustration of the limit case

Note that this is not the maximal violation. The nearest local maximum to this situation appears to be for $A\approx(0.16371,1)$, where the ratio reaches a value of $\frac{MN}{BM}\approx4.17098$. This was found using numeric optimization, so a nice symbolic expression for this is still missing. Chances are that you might be able to prove

$$\frac1{BM}+\frac1{CN}<\frac{4.2}{MN}$$

or something along these lines. To do this, you'd have to show that this local maximum is also the global one. The underlying function, in terms of the $x$ coordinate of both $A$ and $C$, is

sqrt(4*((2*Ax - Cx)*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx^2/(4*(2*Ax - Cx)^2*Ax*Cx^2 -
(2*Ax - Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax -
Cx)^2) - 2)*Cx + sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax +
sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx) -
(2*(2*Ax - Cx)^2*Cx^3 + (2*Ax - Cx)*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax +
sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx^2)/(4*(Ax -
Cx)*(2*Ax - Cx)^2*Cx^2 - (2*Ax - Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 +
4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) -
2)^2*Cx^2))*Cx))^2 + (2*(2*Ax - Cx)*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax +
sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Ax*Cx^2/(4*(2*Ax -
Cx)^2*Ax*Cx^2 - (2*Ax - Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax -
1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx + sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax
- Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) -
2)^2*Cx^2))*Cx) - (2*(Ax - Cx)*(2*Ax - Cx)*(2*(Ax - Cx)*Ax - 2*(2*Ax -
Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx^2 +
(2*Ax - Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax -
Cx)^2) - 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax +
sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) -
2)^2*Cx^2))*Cx^2)/(4*(Ax - Cx)*(2*Ax - Cx)^2*Cx^2 - (2*Ax -
Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2)
- 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx))^2)*(1/sqrt((Cx +
(2*(Ax - Cx)*(2*Ax - Cx)*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax +
sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx^2 + (2*Ax -
Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2)
- 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx^2)/(4*(Ax - Cx)*(2*Ax
- Cx)^2*Cx^2 - (2*Ax - Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax -
1)^2 + 4*(2*Ax - Cx)^2) - 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax
- Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) -
2)^2*Cx^2))*Cx))^2 + 4*(2*(2*Ax - Cx)^2*Cx^3 + (2*Ax - Cx)*(2*(Ax -
Cx)*Ax - 2*(2*Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax -
Cx)^2) - 2)*Cx^2)^2/(4*(Ax - Cx)*(2*Ax - Cx)^2*Cx^2 - (2*Ax -
Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2)
- 2)*Cx - sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx)^2) + 1/sqrt(4*(2*Ax -
Cx)^2*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 +
4*(2*Ax - Cx)^2) - 2)^2*Ax^2*Cx^4/(4*(2*Ax - Cx)^2*Ax*Cx^2 - (2*Ax -
Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2)
- 2)*Cx + sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx)^2 + 4*(2*Ax -
Cx)^2*(2*(Ax - Cx)*Ax - 2*(2*Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 +
4*(2*Ax - Cx)^2) - 2)^2*Cx^4/(4*(2*Ax - Cx)^2*Ax*Cx^2 - (2*Ax -
Cx)*(2*(2*(Ax - Cx)*Ax + sqrt(4*((Ax - Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2)
- 2)*Cx + sqrt(16*(2*Ax - Cx)^2*Cx^2 + 4*(2*(Ax - Cx)*Ax + sqrt(4*((Ax -
Cx)*Ax - 1)^2 + 4*(2*Ax - Cx)^2) - 2)^2*Cx^2))*Cx)^2))

So “all you have to do” is compute the partial derivatives of this beast, solve for them being zero at the same time, enumerate all critical points, and compare them numerically. Good luck finding roots for the partial derivatives, though!

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  • $\begingroup$ oh,Thank you, I have edit $\endgroup$
    – math110
    Aug 7, 2013 at 0:47
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    $\begingroup$ @math110: I must confess that I consider it rather rude to edit a question in such a way that an already given and at that point perfectly valid answer suddenly becomes incorrect. I'm going to try to revert your edit, and if you really want this $4.2$ question answered, post that as a separate question. $\endgroup$
    – MvG
    Aug 7, 2013 at 0:55
  • $\begingroup$ oh,Thank you, very much @MVG $\endgroup$
    – math110
    Aug 7, 2013 at 1:12
  • $\begingroup$ oh,Thank you, very much @MVG,I have post new question: math.stackexchange.com/questions/461510/… $\endgroup$
    – math110
    Aug 7, 2013 at 1:19
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    $\begingroup$ and can you post why $$lim_{\lambda\to\infty}\dfrac{MN}{BM}=\sqrt{\sqrt{4\sqrt{2}+8}+4\sqrt{2}+8}$$ $\endgroup$
    – math110
    Aug 7, 2013 at 1:26
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Since you can scale the triangle without changing the ratio, it is convenient to put $D$ at the origin $(0,0)$ and $A$ at $(-1,0)$. Then your problem looks like this

diagram 1

I can assume $0<\alpha<\phi<\frac{\pi}{2}$.

Then the lines in the figure are given by the following simple formulas

$AC:\quad x\sin\alpha-y\cos\alpha+\sin\alpha=0$,

$AB:\quad x\sin\alpha+y\cos\alpha+\sin\alpha=0$,

$BC:\quad x\sin\phi-y\cos\phi=0$,

$DN:\quad \displaystyle x\sin\frac{\phi+\pi}{2}-y\cos\frac{\phi+\pi}{2}=0$,

$DM:\quad \displaystyle x\sin\frac{\phi}{2}-y\cos\frac{\phi}{2}=0$.

It is now easy to find the coordinates of the points by intersecting the lines, which gives:

$\displaystyle N=\left(\frac{\tan\alpha}{\tan(\phi+\pi)/2-\tan\alpha}, \frac{\tan\alpha\tan(\phi+\pi)/2}{\tan(\phi+\pi)/2-\tan\alpha}\right)$,

$M=\displaystyle\left(-\frac{\tan\alpha}{\tan\phi/2+\tan\alpha}, -\frac{\tan\alpha\tan\phi/2}{\tan\phi/2+\tan\alpha}\right)$,

$B=\displaystyle\left(-\frac{\tan\alpha}{\tan\phi+\tan\alpha}, -\frac{\tan\alpha\tan\phi}{\tan\phi+\tan\alpha}\right)$,

$C=\displaystyle\left(\frac{\tan\alpha}{\tan\phi-\tan\alpha}, \frac{\tan\alpha\tan\phi}{\tan\phi-\tan\alpha}\right)$,

Plugging these into the usual Euclidean distance formula gives $MN$, $BM$ and $CN$. After some simplification with mathematica, I get the following

$MN=\displaystyle\frac{2\sin\alpha\sqrt{1+\sin\phi\sin2\alpha}}{\sin\phi+\sin2\alpha}$,

$BM=\displaystyle\frac{\sin\alpha\sin\phi/2}{\sin(\phi+\alpha)\sin(\phi/2+\alpha)}$,

$CN=\displaystyle\frac{\sin\alpha\cos\phi/2}{\sin(\phi-\alpha)\cos(\phi/2-\alpha)}$.

Then $R(\alpha,\phi)=\displaystyle\frac{MN}{BM}+\frac{MN}{CN}$ is a function of two variables $\alpha,\phi$ and it looks like this

diagram 2

And here is another plot showing just the bit that sticks out above 4

diagram 3

It is obvious from these plots that the maximum of $R(\alpha,\phi)$ occurs on the diagonal $\alpha=\phi$, where

$$r(\alpha)=R(\alpha,\alpha)= 2\sqrt{2}\cos\alpha \sqrt{2 + \cos\alpha - \cos3\alpha}.$$

It reaches the maximum at $\alpha_m$ which satisfies $-8 \sin\alpha_m - 2 \sin2\alpha_m + 5 \sin4\alpha_m=0$. This equation can be solved in mathematica analytically but the result is pretty ugly. For numerical values, I get $\alpha_m=0.4371582260685195$ and $r(\alpha_m)=4.170978697039997$. So, the correct inequality is

$$\frac{1}{BM}+\frac{1}{CN}\le\frac{r(\alpha_m)}{MN}$$

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    $\begingroup$ $$r(\alpha_m) = \frac{4}{5}\sqrt{\frac {\sqrt[3]{2625000 \sqrt{5} + 146875000} + \sqrt[3]{3331125 \sqrt{5} + 93341025} + 123 \sqrt[3]{50}} {5\sqrt[3]{21\sqrt{5} + 1175}}}$$ satisfies $$3125\,r(\alpha_m)^6 - 60000\,r(\alpha_m)^4 + 100608\,r(\alpha_m)^2 - 45056$$ which is a condition I obtained from your equation. Nice! $\endgroup$
    – MvG
    Aug 7, 2013 at 13:58
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    $\begingroup$ For completness' sake: $$\alpha_m=\arccos\left(\sqrt[3]{\frac{1}{10\sqrt5}+\frac{1}{10}} + \frac{1}{5\sqrt[3]{\frac{1}{10\sqrt5}+\frac{1}{10}}}\right)$$ satisfying $$5\cos^3\alpha_m - 3\cos\alpha_m - 1 = 0\;.$$ Note that “it is obvious from these plots” does not constitute a proof, strictly speaking. $\endgroup$
    – MvG
    Aug 7, 2013 at 14:26
  • $\begingroup$ Honestly, I was surprised to get as far as I did. The function $R(\alpha,\phi)$ is fairly well behaved, so from the applied math point of view I would be inclined to trust my eyes more than a formal proof. $\endgroup$ Aug 9, 2013 at 4:48
  • $\begingroup$ If you are concerned about the limit of $R$ at $(\alpha,\phi)\to(0,0)$, then you might want to consider $S(k,\phi)=R(k\phi,\phi)$, where $0<k<1$. The plot of $S$ looks very smooth, so I don't expect any nasty surprises from it. $\endgroup$ Aug 9, 2013 at 4:52
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Hint:We consider $\Delta ABC$ as triangle in $\mathbb{R}^2$($O$ origin is midpoint of $BC$)and encircling it in the ellipse with the property that its two foci are $B(-a,0),C(a,0)$ and passes $A(x_0,y_0)$.

canonical equation of the ellipse is $\frac{x^2}{d^2}$+${y^2}\frac{1-\frac{x_0^2}{d^2}}{y_0^2}=1$,$d=\frac{1}{2}(\sqrt{(x_0-a)^2+y_0^2}+\sqrt{(x_0+a)^2+y_0^2})$, then the normal line to tangent line of the ellipse at $A$ is bisector of $\angle BAC$(here) now we find Coordinates of $D$,similarly we find $M$ and $N$,now compute $BM,CN,MN$

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