In a non right triangle $PQR$, the median from $R$ meets the sides $PQ$ at $S$, the perpendicular from $P$ meets sides $QR$ at $E$ and $RS$ and $PE$ intersect at $O$. $p=\sqrt{3},q=1$ and circumradius of $PQR$ is $1$.

Aganippe76
2022-07-11
Answered

In a non right triangle $PQR$, the median from $R$ meets the sides $PQ$ at $S$, the perpendicular from $P$ meets sides $QR$ at $E$ and $RS$ and $PE$ intersect at $O$. $p=\sqrt{3},q=1$ and circumradius of $PQR$ is $1$.

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Antonio Dickerson

Answered 2022-07-12
Author has **5** answers

Note that $PE$ is a median and hence $PE=3OE$. So $\mathrm{\u25b3}PQR=3\times \mathrm{\u25b3}OQR$.

The question states that $\mathrm{\u25b3}PQR$ is not a right angled triangle otherwise $\mathrm{\angle}R={90}^{0}$ is another obvious configuration for the given sides and circumradius.

As $p=\sqrt{3}>q=1,\mathrm{\angle}Q\phantom{\rule{thinmathspace}{0ex}}$ is acute. That leads to only other possible configuration of $\mathrm{\angle}P\phantom{\rule{thinmathspace}{0ex}}$ being obtuse.

If $C$ is the circumcenter, you can see why $\mathrm{\angle}P={120}^{0}\phantom{\rule{thinmathspace}{0ex}}PQ=PR$ and $E$ is the midpoint of $QR$.

The question states that $\mathrm{\u25b3}PQR$ is not a right angled triangle otherwise $\mathrm{\angle}R={90}^{0}$ is another obvious configuration for the given sides and circumradius.

As $p=\sqrt{3}>q=1,\mathrm{\angle}Q\phantom{\rule{thinmathspace}{0ex}}$ is acute. That leads to only other possible configuration of $\mathrm{\angle}P\phantom{\rule{thinmathspace}{0ex}}$ being obtuse.

If $C$ is the circumcenter, you can see why $\mathrm{\angle}P={120}^{0}\phantom{\rule{thinmathspace}{0ex}}PQ=PR$ and $E$ is the midpoint of $QR$.

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