In-a-equilateral-triangle-ABC-whose-side-is-a-the-points-M-and-N-are-taken-on-the-side-BC-such-that-the-triangles-ABM-AMN-and-ANC-have-the-same-perimeter-Calculate-the-distances-from-vertex-A-to-

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Question Number 68664 by Maclaurin Stickker last updated on 14/Sep/19

$$\mathrm{In}\mathrm{a}\mathrm{equilateral}\mathrm{triangle}ABC\mathrm{whose}$$$$\mathrm{side}\mathrm{is}\mathit{a},\mathrm{the}\mathrm{points}M\mathrm{and}N\mathrm{are}\mathrm{taken}$$$$\mathrm{on}\mathrm{the}\mathrm{side}BC,\mathrm{such}\mathrm{that}\mathrm{the}\mathrm{triangles}$$$$ABM,AMN\mathrm{and}ANC\mathrm{have}\mathrm{the}\mathrm{same}$$$$\mathrm{perimeter}.\mathrm{Calculate}\mathrm{the}\mathrm{distances}\mathrm{from}$$$$\mathrm{vertex}A\mathrm{to}\mathrm{points}M\mathrm{and}N.$$$$(\mathbf{solve}\mathbf{in}\mathbf{detail}.)$$

Answered by mr W last updated on 14/Sep/19

Commented by mr W last updated on 14/Sep/19

$$letBM=NC=x$$$$\Rightarrow MN=a-2x$$$$AM=AN=\sqrt{a^2+x^2-2ax\cos 60°}=\sqrt{a^2+x^2-ax}$$$$a+x+AM=2AM+MN$$$$a+x=AM+a-2x$$$$\Rightarrow AM=3x$$$$\sqrt{a^2+x^2-ax}=3x$$$$a^2+x^2-ax=9x^2$$$$8x^2+ax-a^2=0$$$$x=\frac{(\sqrt{33}-1)a}{16}\approx 0.2965a$$$$\Rightarrow AM=AN=3x=\frac{3(\sqrt{33}-1)a}{16}\approx 0.8796a$$

Commented by Rasheed.Sindhi last updated on 15/Sep/19

$$\mathbf{Sir},\mathrm{why}\mathrm{have}\mathrm{you}\mathrm{assumed}BM=NC?$$

Commented by mr W last updated on 15/Sep/19

$$theassumptionisbasedonsymmetry.$$$$certainlyitcanalsobeprovedthat$$$$BMmustbeequaltoNC.$$

Commented by mr W last updated on 15/Sep/19

$$letBM=x,NC=y\ne x$$$$a+x+\sqrt{x^2+a^2-ax}=a-x-y+\sqrt{x^2+a^2-ax}+\sqrt{y^2+a^2-ay}$$$$\Rightarrow \sqrt{y^2+a^2-ay}=2x+y$$$$\Rightarrow a^2-ay=4x^2+4xy\dots \left(i\right)$$$$$$$$a+x+\sqrt{x^2+a^2-ax}=a+y+\sqrt{y^2+a^2-ay}$$$$\Rightarrow x+\sqrt{x^2+a^2-ax}=y+\sqrt{y^2+a^2-ay}$$$$\Rightarrow x+\sqrt{x^2+a^2-ax}=y+2x+y$$$$\Rightarrow \sqrt{x^2+a^2-ax}=2y+x$$$$\Rightarrow a^2-ax=4y^2+4xy\dots \left(ii\right)$$$$\left(i\right)-\left(ii\right):$$$$a(x-y)=4(x^2-y^2)$$$$\Rightarrow a(x-y)=4(x-y)(x+y)$$$$\Rightarrow x-y=0ora=4(x+y)$$$$\Rightarrow x=yorx+y=\frac{a}{4}$$$$butifx+y=\frac{a}{4}:$$$$\left(i\right)+\left(ii\right):$$$$2a^2-a(x+y)=4(x^2+y^2+2xy)$$$$2a^2-a(x+y)=4{(x+y)}^2$$$$2a^2-\frac{a^2}{4}=\frac{a^2}{4}$$$$\frac{3a^2}{2}=0!$$$$\Rightarrow x=yistheonlypossibility!$$

Commented by Rasheed.Sindhi last updated on 15/Sep/19

$$\mathcal{T}h\alpha n\mathcal{X}alotsir.{You}^{\prime }\mathrm{re}verydeep$$$$ingeometryalso!$$

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