MathLabs

Problem 1

Let ABCABC be an acute triangle inscribed in a circle Γ\Gamma. Let A1A_1 be the orthogonal projection of AA onto BCBC, so that AA1AA_1 is an altitude. Let B1B_1 and C1C_1 be the orthogonal projections of A1A_1 onto ABAB and ACAC, respectively. Point PP is such that quadrilateral AB1PC1AB_1PC_1 is convex and has the same area as triangle ABCABC. Is it possible that PP lies strictly in the interior of circle Γ\Gamma? Justify your answer.
Step 3 of 5: AB1DC1 already has the right area
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[AB1DC1]=B1C1⋅AD2=AA1⋅BC2=[ABC][AB_1DC_1]=\frac{B_1C_1\cdot AD}{2}=\frac{AA_1\cdot BC}{2}=[ABC]
Detailed analysis

Write AD=2RAD=2R; the law of sines gives BC=2Rsin⁡ABC=2R\sin A. Since B1C1=AA1sin⁡AB_1C_1=AA_1\sin A (Step 1) and AD⊥B1C1AD\perp B_1C_1 (Step 2), the area of the convex quadrilateral AB1DC1AB_1DC_1 equals B1C1⋅AD2=AA1sin⁡A⋅2R2=AA1⋅BC2\frac{B_1C_1\cdot AD}{2}=\frac{AA_1\sin A\cdot 2R}{2}=\frac{AA_1\cdot BC}{2}, which is exactly the area of triangle ABCABC computed from base BCBC and height AA1AA_1.