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Geometry Difficulty 5.6 AIME, harder Find the answer

Example 11 As shown in Figure 2.10.10, through any point PP inside △ABC\triangle ABC, draw three lines parallel to its sides, forming three triangles with two of these lines and one side of the triangle. Let the areas of these three triangles be S1,S2,S3S_{1}, S_{2}, S_{3}. Try to find such a point that S1+S2+S3S_{1}+S_{2}+S_{3} reaches its minimum value.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let AB=c,DP=AL=u,GP=BK=vA B=c, D P=A L=u, G P=B K=v, then LK=c−u−vL K=c-u-v. Also,
S△DPE=u2⋅sin⁡A⋅sin⁡B2sin⁡2C=S1S△FμG=v2⋅sin⁡A⋅sin⁡B2sin⁡2C=S2S△PLX=(c−u−v)2⋅sin⁡A⋅sin⁡B2sin⁡2C=S3 \begin{array}{c} S_{\triangle D P E}=\frac{u^{2} \cdot \sin A \cdot \sin B}{2 \sin ^{2} C}=S_{1} \\ S_{\triangle F \mu G}=\frac{v^{2} \cdot \sin A \cdot \sin B}{2 \sin ^{2} C}=S_{2} \\ S_{\triangle P L X}=\frac{(c-u-v)^{2} \cdot \sin A \cdot \sin B}{2 \sin ^{2} C}=S_{3} \end{array}

Thus,
S1+S2+S3=sin⁡A⋅sin⁡B2sin⁡2C[u2+v2+(c−u−v)2]⩾3sin⁡A⋅sin⁡B2sin⁡2C[u+v+(c−u−v)3]2 (Quadratic mean and arithmetic mean inequality) =c2sin⁡A⋅sin⁡B6sin⁡2C \begin{aligned} S_{1}+S_{2}+S_{3}= & \frac{\sin A \cdot \sin B}{2 \sin ^{2} C}\left[u^{2}+v^{2}+(c-u-v)^{2}\right] \geqslant \\ & \frac{3 \sin A \cdot \sin B}{2 \sin ^{2} C}\left[\frac{u+v+(c-u-v)}{3}\right]^{2} \\ & \text { (Quadratic mean and arithmetic mean inequality) }= \\ & \frac{c^{2} \sin A \cdot \sin B}{6 \sin ^{2} C} \end{aligned}

The equality holds if and only if u=v=c−u−v=c3u=v=c-u-v=\frac{c}{3}.
Therefore, when PP is the centroid of △ABC\triangle A B C, S1+S2+S3S_{1}+S_{2}+S_{3} achieves its minimum value.

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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.