Question Number 16803 by ajfour last updated on 26/Jun/17 Commented by ajfour last updated on 26/Jun/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{in}\:\mathrm{an}\:\mathrm{equilateral}\: \\ $$$$\mathrm{triangle}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{distances} \\ $$$$\mathrm{from}\:\mathrm{any}\:\mathrm{point}\:\mathrm{to}\:\mathrm{the}\:\mathrm{three}\:\mathrm{sides} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{constant}. \\ $$…
Question Number 16771 by Tinkutara last updated on 26/Jun/17 Commented by Tinkutara last updated on 26/Jun/17 $$\left(\mathrm{1}\right)\:{O}\:\mathrm{is}\:\mathrm{circumcenter}\:\mathrm{and}\:{H}\:\mathrm{is}\:\mathrm{orthocenter} \\ $$$$\mathrm{of}\:\Delta{ABC}.\:{AOA}'\:\mathrm{is}\:\mathrm{the}\:\mathrm{diameter}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:{A}'{BHC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{parallelogram}. \\ $$ Answered by…
Question Number 82308 by Power last updated on 20/Feb/20 Commented by mr W last updated on 20/Feb/20 $$\left.{D}\right)\:\mathrm{25} \\ $$ Terms of Service Privacy Policy…
Question Number 16785 by RasheedSoomro last updated on 26/Jun/17 Commented by RasheedSoomro last updated on 26/Jun/17 $$\mathrm{Without}\:\mathrm{using}\:\mathrm{area}-\mathrm{formula}\:\mathrm{find}\:\mathrm{out} \\ $$$$\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{shaded}\:\mathrm{part}\:\mathrm{to}\:\mathrm{the}\:\mathrm{whole}. \\ $$ Commented by mrW1 last…
Question Number 82302 by Power last updated on 20/Feb/20 Commented by mr W last updated on 20/Feb/20 $${x}=\mathrm{20}° \\ $$$${see}\:{Q}\mathrm{61347} \\ $$ Commented by Power…
Question Number 16756 by Tinkutara last updated on 26/Jun/17 $$\mathrm{Alternate}\:\mathrm{vertices}\:\mathrm{of}\:\mathrm{a}\:\mathrm{regular}\:\mathrm{hexagon} \\ $$$$\mathrm{are}\:\mathrm{joined}\:\mathrm{as}\:\mathrm{shown}.\:\mathrm{What}\:\mathrm{fraction}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{total}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hexagon}\:\mathrm{is} \\ $$$$\mathrm{shaded}?\:\left(\mathrm{Justify}\:\mathrm{your}\:\mathrm{answer}.\right) \\ $$ Commented by Tinkutara last updated on 26/Jun/17…
Question Number 16754 by ajfour last updated on 26/Jun/17 Commented by ajfour last updated on 26/Jun/17 $$\mathrm{Q}.\mathrm{16748}\:\left(\mathrm{solution}\right) \\ $$$$\:\mathrm{by}\:\mathrm{fault}\:\mathrm{it}\:\mathrm{gets}\:\mathrm{uploaded}\:\mathrm{as}\: \\ $$$$\mathrm{question}. \\ $$ Answered by…
Question Number 16740 by Tinkutara last updated on 26/Jun/17 $$\mathrm{The}\:\mathrm{maximum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\left(\mathrm{cos}\:\left(\mathrm{33}\pi\:+\:\theta\right)\right)\:+\:\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{sin}\:\left(\mathrm{45}\pi\:+\:\theta\right)\right) \\ $$$$\mathrm{is} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{1}\:+\:\mathrm{sin}^{\mathrm{2}} \mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{2} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{1}\:+\:\mathrm{cos}^{\mathrm{2}} \mathrm{1} \\…
Question Number 16738 by Tinkutara last updated on 26/Jun/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{segments}\:\mathrm{joining}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{opposite}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{equiangular}\:\mathrm{hexagon}\:\mathrm{are}\:\mathrm{concurrent}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 16739 by Tinkutara last updated on 26/Jun/17 $$\mathrm{Let}\:{M}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interior}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}\:\mathrm{and}\:\mathrm{let}\:{A}', \\ $$$${B}'\:\mathrm{and}\:{C}'\:\mathrm{be}\:\mathrm{its}\:\mathrm{projections}\:\mathrm{onto}\:\mathrm{the} \\ $$$$\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB},\:\mathrm{respectively}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{inradii}\:\mathrm{of}\:\mathrm{triangles}\:{MAC}',\:{MBA}'\:\mathrm{and} \\ $$$${MCB}'\:\mathrm{equals}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{inradii}\:\mathrm{of}\:\mathrm{trianges}\:{MAB}',\:{MBC}'\:\mathrm{and} \\…