in the triangle ABC on the AC side, points M and N are chosen such that ABM = MBN = NBC

Question

in the triangle ABC on the AC side, points M and N are chosen such that <ABM = <MBN = <NBC It turned out that NB = BC. On the side AB, a point K was marked such that BK = BM. Prove that AK + NC> AM.

I tried to get a triangle with sides AK NC and AM. But I couldn't. So I don't know how to prove this

Answer 1

let $BC=a$ and $\measuredangle ABC=3\beta$.

Thus, $$\measuredangle ACB=90^{\circ}-\frac{\beta}{2},$$ $$\measuredangle ANC=90^{\circ}+\frac{\beta}{2},$$ $$\measuredangle AMB=90^{\circ}+\frac{3\beta}{2}$$ and $$\measuredangle BAC=180^{\circ}-\left(90^{\circ}+\frac{5\beta}{2}\right),$$ which says $$90^{\circ}+\frac{5\beta}{2}<180^{\circ},$$ which gives $$0^{\circ}<\beta<36^{\circ}.$$ Now, $$NC=2a\sin\frac{\beta}{2}$$ and by law of sines we can show that: $$AM=\frac{a\cos\frac{\beta}{2}\sin\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ and $$AK=AB-BK=\frac{a\cos\frac{\beta}{2}}{\cos\frac{5\beta}{2}}-\frac{a\cos\frac{\beta}{2}}{\cos\frac{3\beta}{2}}=\frac{a\sin\beta\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}.$$ Id est, we need to prove that: $$\frac{a\sin\beta\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}+2a\sin\frac{\beta}{2}>\frac{a\cos\frac{\beta}{2}\sin\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ or $$\frac{\cos\frac{\beta}{2}\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}+1>\frac{\cos^2\frac{\beta}{2}}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ or $$2\cos\frac{\beta}{2}\sin2\beta+\cos4\beta+\cos\beta>1+\cos\beta$$ or $$\cos\frac{\beta}{2}\sin2\beta>2\sin^22\beta$$ or $$1>4\sin\frac{\beta}{2}\cos\beta$$ or $$1>4\sin\frac{\beta}{2}\left(1-2\sin^2\frac{\beta}{2}\right)$$ or $$1-2\sin\frac{\beta}{2}>2\sin\frac{\beta}{2}\left(1-4\sin^2\frac{\beta}{2}\right)$$ $$1>2\sin\frac{\beta}{2}\left(1+2\sin\frac{\beta}{2}\right)$$ or $$\sin\frac{\beta}{2}<\frac{\sqrt5-1}{4},$$ which we got by looking for domain of $\beta$.

Answer 2

Geometric hint: Consider the figure:

Following cases can be considered:

a-If BC=BF then triangle BCF is isosceles then NC=DE and FD+DE>FE. This is a particular case.

b-If BC<AB then we have:

AK+KG=AK+DE>AM

Since $\angle AKM>90^o$ and:

$NC>DE>\frac{KM}{2}$

Therefore:

AK+NC>AM

Answer 3

Let $|CH|=|HN|$, so $BH\perp AC$ and $|BH|=h$ is the height of $\triangle ABC$, $\triangle MBC$, $\triangle NBC$. Also let $|BC|=a=|BN|$, $|BM|=d=|BK|$.

\begin{align} a&=\frac h{\cos\phi} ,\\ |NC|&=2h\tan\phi .\\ d&=\frac h{\cos 3\phi} ,\\ |AK|&= \frac h{\cos 5\phi}-d = h\cdot\frac{\cos 3\phi-\cos 5\phi}{\cos 3\phi \cos 5\phi} ,\\ |AM|&= h\,(\tan 5\phi-\tan 3\phi) \\ &= \frac{8h\sin^2\phi\cos 2\phi}{\cos\phi(2\sin\phi+1)} \cdot \frac 1{1+2\sin\phi-4\sin^2\phi} ,\\ \end{align}

Since $\sin\phi,\cos\phi,\cos2\phi$ are positive for $\phi\in(0,\tfrac\pi6)$ the question simplifies to

\begin{align} 1+2\sin\phi-4\sin^2\phi &>0 \quad \text{for } \phi\in(0.\tfrac\pi3) . \end{align}

Answer 4

Hint: One can prove that NC>KM, and AK+KM>AM thus the requirement will be satisfied.

(':=' was used as denoting)

we have: $$<NBC=<MBN:=α,$$ we also have that $$BN=BC =>$$ $$<BNC=<BCN = {\frac{180-α}{2}}:=β$$ its evident that β<90 (since 2β<180 in triangle BCN) => $$<MBN > 90, $$$$<MNB:=β+x$$ and since $$<MBN=α =>$$$$ <BMN=180-α-(β+x)=β-x<β =>$$$$ BM>BN$$ (is located in front of bigger angle) $$=>$$ we have to triangles BKM, BNC with the same angle α but edges of the first are bigger than the edges of the second => base of the first < base of the second $$=> KM<NC,$$ and since $$AM+KM>AM => $$$$AM+NC>AM$$

P.S I would kindly ask someone to edit/improve this solution.