Question Number 19415 by Tinkutara last updated on 10/Aug/17 $${PS}\:\mathrm{is}\:\mathrm{a}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{of}\:\mathrm{length}\:\mathrm{4}\:\mathrm{and}\:{O} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:{PS}.\:\mathrm{A}\:\mathrm{semicircular} \\ $$$$\mathrm{arc}\:\mathrm{is}\:\mathrm{drawn}\:\mathrm{with}\:{PS}\:\mathrm{as}\:\mathrm{diameter}.\:\mathrm{Let} \\ $$$${X}\:\mathrm{be}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{this}\:\mathrm{arc}.\:{Q}\:\mathrm{and}\:{R} \\ $$$$\mathrm{are}\:\mathrm{points}\:\mathrm{on}\:\mathrm{the}\:\mathrm{arc}\:{PXS}\:\mathrm{such}\:\mathrm{that}\:{QR} \\ $$$$\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{PS}\:\mathrm{and}\:\mathrm{the}\:\mathrm{semicircular} \\ $$$$\mathrm{arc}\:\mathrm{drawn}\:\mathrm{with}\:{QR}\:\mathrm{as}\:\mathrm{diameter}\:\mathrm{is} \\ $$$$\mathrm{tangent}\:\mathrm{to}\:{PS}.\:\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the} \\…
Question Number 19394 by tawa tawa last updated on 10/Aug/17 Answered by allizzwell23 last updated on 10/Aug/17 $$ \\ $$$$\:\:\:\frac{\mathrm{AB}}{\mathrm{DF}}\:=\:\frac{\mathrm{AC}}{\mathrm{CE}}\:\:\mathrm{similar}\:\mathrm{triangles} \\ $$$$\:\:\mathrm{Let}\:\mathrm{AD}\:=\:\mathrm{x}\:\:\Rightarrow\:\mathrm{BF}\:=\:\mathrm{x}\:\:\:\therefore\:\mathrm{AB}\:=\:\mathrm{6}+\mathrm{2x} \\ $$$$\:\:\frac{\mathrm{2x}+\mathrm{6}}{\mathrm{6}}\:=\:\frac{\mathrm{20}}{\mathrm{12}}\:\:\:\:\Rightarrow\:\mathrm{2x}+\mathrm{6}\:=\:\frac{\mathrm{20}}{\mathrm{12}}×\mathrm{6}\:=\:\mathrm{10} \\…
Question Number 19388 by mrW1 last updated on 10/Aug/17 $$\mathrm{related}\:\mathrm{to}\:\mathrm{Q}.\mathrm{19333} \\ $$$$\mathrm{the}\:\mathrm{side}\:\mathrm{lengthes}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{are}\: \\ $$$$\mathrm{integer}.\:\mathrm{if}\:\mathrm{the}\:\mathrm{perimeter}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle} \\ $$$$\mathrm{is}\:\mathrm{100},\:\mathrm{how}\:\mathrm{many}\:\mathrm{different}\:\mathrm{triangles} \\ $$$$\mathrm{exist}?\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{of} \\ $$$$\mathrm{them}? \\ $$ Commented by mrW1…
Question Number 150451 by ajfour last updated on 12/Aug/21 Commented by Ar Brandon last updated on 12/Aug/21 $$\mathrm{I}\:\mathrm{admire}\:\mathrm{your}\:\mathrm{posts}\:\mathrm{sir}.\:\mathrm{Sadly}\:\mathrm{I}\:\mathrm{have}\:\mathrm{very}\:\mathrm{little}\:\mathrm{knowledge} \\ $$$$\mathrm{on}\:\mathrm{geometry}.\:\mathrm{You}'\mathrm{re}\:\mathrm{so}\:\mathrm{advanced}.\:\mathrm{Hahaha}\:!\:\mathrm{I}\:\mathrm{hope}\:\mathrm{I}\:\mathrm{will}\:\mathrm{get} \\ $$$$\mathrm{there}\:\mathrm{someday}. \\ $$ Commented…
Question Number 84899 by Power last updated on 17/Mar/20 Commented by Tony Lin last updated on 17/Mar/20 $${A}\left({ADE}\right)=\mathrm{9} \\ $$$$\Rightarrow{h}=\mathrm{6} \\ $$$${A}\left({EFC}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{EC}×\left({h}×\frac{\mathrm{6}}{\mathrm{7}}\right) \\…
Question Number 84892 by Power last updated on 17/Mar/20 Commented by Power last updated on 17/Mar/20 $$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{see}} \\ $$ Commented by Power last updated on…
Question Number 150418 by ajfour last updated on 12/Aug/21 Commented by ajfour last updated on 12/Aug/21 $${For}\:{tetrahedron}\:{edge}\:{a}, \\ $$$${and}\:{sphere}\:{radius}\:{r},\:{find}\: \\ $$$${fractional}\:{volume}\:{of}\:{sphere}\: \\ $$$${within}\:{the}\:{tetrahedron}. \\ $$$${f}=\begin{cases}{{g}_{\mathrm{1}}…
Question Number 150376 by cherokeesay last updated on 11/Aug/21 Commented by liberty last updated on 12/Aug/21 $$\left(\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{r}\right)^{\mathrm{2}} =\left(\mathrm{4}\sqrt{\mathrm{3}}−\mathrm{r}\right)^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{r}\right)^{\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{2}\sqrt{\mathrm{3}}\:=\:{a} \\ $$$$\Rightarrow\left({a}+{r}\right)^{\mathrm{2}} −\left({a}−{r}\right)^{\mathrm{2}} =\left(\mathrm{2}{a}−{r}\right)^{\mathrm{2}}…
Question Number 19301 by saa last updated on 09/Aug/17 $${e}^{{i}\pi} +\mathrm{1}=\mathrm{0} \\ $$ Answered by ajfour last updated on 09/Aug/17 $$\mathrm{e}^{\mathrm{i}\pi} =−\mathrm{1}\:. \\ $$ Terms…
Question Number 19293 by Tinkutara last updated on 08/Aug/17 $$\mathrm{Let}\:{AC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{and}\:{B}\:\mathrm{a}\:\mathrm{point}\:\mathrm{between}\:{A}\:\mathrm{and}\:{C}. \\ $$$$\mathrm{Construct}\:\mathrm{isosceles}\:\mathrm{triangles}\:{PAB}\:\mathrm{and} \\ $$$${QBC}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{segment}\:{AC} \\ $$$$\mathrm{such}\:\mathrm{that}\:\angle{APB}\:=\:\angle{BQC}\:=\:\mathrm{120}°\:\mathrm{and} \\ $$$$\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle}\:{RAC}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{side}\:\mathrm{of}\:{AC}\:\mathrm{such}\:\mathrm{that}\:\angle{ARC}\:=\:\mathrm{120}°. \\ $$$$\mathrm{Show}\:\mathrm{that}\:{PQR}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral} \\…