<APM=? BQ=AC , CQ=2PM, midpoint (2021 All-Siberian Qualifying 10.4)

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<APM=? BQ=AC , CQ=2PM, midpoint (2021 All-Siberian Qualifying 10.4)

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#1 h May 18, 2021, 11:04 PM

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Point $M$ is the midpoint of side $AB$ of triangle $ABC$ . On $CM$ lie points $P$ and $Q$ so that $P$ is closer to $M$ , $Q$ is closer to $C$ , and $CQ = 2PM$ . It turned out that $BQ = AC$ . Find the measuere of the angle $APM$ .

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#2 h May 21, 2021, 12:08 PM

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Given $A(-a,0),B(b,0),C(0,c)$ .

Midpoint $M(\frac{b-a}{2},0)$ , equation of the line $CM\ :\ y=\frac{2c}{a-b}(x-\frac{b-a}{2})$ .
Point $P \in CM\ :\ P(\lambda,\frac{2c}{a-b}(\lambda-\frac{b-a}{2})\ )$ .

Condition $CQ=2 \cdot PM$ gives the point $Q(b-a-2\lambda,\frac{2c}{a-b}(\frac{b-a}{2}-2\lambda)\ )$ .
Second condition $\Rightarrow \lambda=-\frac{(a-b)(a^{2}-ab+2c^{2})}{(a-b)^{2}+4c^{2}}$ .

Slope of the line $AP\ :\ m_{AP}=\frac{b-a}{2c}$ .
Because the slope of the line $CM\ :\ m_{CM}=\frac{2c}{a-b}$ , we find $CM \bot AP$ .

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#3 h May 21, 2021, 4:42 PM

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My solution at https://stanfulger.blogspot.com/2021/05/2021-all-siberian-qualifying-form-10.html

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sunken rock

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