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Geometry Difficulty 5.6 AIME, harder Prove it

Example 1. In square ABCD\mathrm{ABCD}, M\mathrm{M} is on CD\mathrm{CD}, and extending AM\mathrm{AM} intersects BC\mathrm{BC} at N\mathrm{N} (Figure 1).

Prove: AM+AN>2AC\mathrm{AM}+\mathrm{AN} > 2 \mathrm{AC}.

Solution

Prove that for a square with side length aa, ∠CAN=α\angle \mathrm{CAN}=\alpha, then
∠N=∠DAN=45∘−α,AC=2a,AM=acos⁡(45∘−α). \begin{array}{l} \angle \mathrm{N}=\angle \mathrm{DAN} \\ =45^{\circ}-\alpha, \\ \mathrm{AC}=\sqrt{2} \mathrm{a}, \\ \mathrm{AM}=\frac{\mathrm{a}}{\cos \left(45^{\circ}-\alpha\right)}. \end{array}

In △ACN\triangle \mathrm{ACN}, ∠ACN=135∘\angle \mathrm{ACN}=135^{\circ},
AN=2asin⁡135∘sin⁡(45∘−α)=asin⁡(45∘−α). Therefore, AM+AN=a[1cos⁡(45∘−α)+1sin⁡(45∘−α)]=22cos⁡αcos⁡2α.∵0∘<α<45∘, Therefore, AM+AN>22a=2AC. \begin{array}{l} \mathrm{AN}=\frac{\sqrt{2} a \sin 135^{\circ}}{\sin \left(45^{\circ}-\alpha\right)}=\frac{a}{\sin \left(45^{\circ}-\alpha\right)}. \\ \text { Therefore, } \mathrm{AM}+\mathrm{AN}=\mathrm{a}\left[\frac{1}{\cos \left(45^{\circ}-\alpha\right)}\right. \\ \left.+\frac{1}{\sin \left(45^{\circ}-\alpha\right)}\right]=\frac{2 \sqrt{2} \cos \alpha}{\cos 2 \alpha}. \\ \because 0^{\circ}<\alpha<45^{\circ}, \\ \text { Therefore, } \mathrm{AM}+\mathrm{AN}>2 \sqrt{2} \mathrm{a}=2 \mathrm{AC}. \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.