Question Number 85352 by Power last updated on 21/Mar/20 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 19792 by Tinkutara last updated on 15/Aug/17 $$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{with}\:\angle{DAB}\:=\:\angle{BDC}\:=\:\mathrm{90}°.\:\mathrm{Let}\:\mathrm{the} \\ $$$$\mathrm{incircles}\:\mathrm{of}\:\mathrm{triangles}\:{ABD}\:\mathrm{and}\:{BCD} \\ $$$$\mathrm{touch}\:{BD}\:\mathrm{at}\:{P}\:\mathrm{and}\:{Q},\:\mathrm{respectively}, \\ $$$$\mathrm{with}\:{P}\:\mathrm{lying}\:\mathrm{in}\:\mathrm{between}\:{B}\:\mathrm{and}\:{Q}.\:\mathrm{If} \\ $$$${AD}\:=\:\mathrm{999}\:\mathrm{and}\:{PQ}\:=\:\mathrm{200}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is} \\ $$$$\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{radii}\:\mathrm{of}\:\mathrm{the}\:\mathrm{incircles}\:\mathrm{of} \\ $$$$\mathrm{triangles}\:{ABD}\:\mathrm{and}\:{BDC}? \\…
Question Number 19794 by Tinkutara last updated on 15/Aug/17 $$\mathrm{In}\:\mathrm{a}\:\mathrm{triangle}\:{ABC}\:\mathrm{with}\:\angle{BCA}\:=\:\mathrm{90}°, \\ $$$$\mathrm{the}\:\mathrm{perpendicular}\:\mathrm{bisector}\:\mathrm{of}\:{AB} \\ $$$$\mathrm{intersects}\:\mathrm{segments}\:{AB}\:\mathrm{and}\:{AC}\:\mathrm{at}\:{X} \\ $$$$\mathrm{and}\:{Y},\:\mathrm{respectively}.\:\mathrm{If}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{quadrilateral}\:{BXYC}\:\mathrm{to}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{triangle}\:{ABC}\:\mathrm{is}\:\mathrm{13}\::\:\mathrm{18}\:\mathrm{and} \\ $$$${BC}\:=\:\mathrm{12}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of}\:{AC}? \\ $$ Terms…
Question Number 19783 by NECC last updated on 15/Aug/17 $${The}\:{sides}\:{of}\:{a}\:{triangle}\:{are}\:{of} \\ $$$${lengths}\:\sqrt{\left({m}^{\mathrm{2}} −{n}^{\mathrm{2}} \right)}\:,{m}^{\mathrm{2}} +{n}^{\mathrm{2}} ,\:\mathrm{2}{mn}. \\ $$$${Show}\:{that}\:{it}\:{is}\:{a}\:{right}\:{angle}\:\Delta. \\ $$ Commented by Tinkutara last updated…
Question Number 85298 by mr W last updated on 20/Mar/20 $${Find}\:{the}\:{symmetry}\:{axes}\:\left({if}\:{any}\right)\:{of} \\ $$$${the}\:{following}\:{curve}: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{2}{x}−\mathrm{8}{y}−\mathrm{2}=\mathrm{0} \\ $$ Commented by jagoll last updated on…
Question Number 150789 by cherokeesay last updated on 15/Aug/21 Answered by nimnim last updated on 15/Aug/21 $$\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\mathrm{1}}{\:\sqrt{{R}}}+\frac{\mathrm{1}}{\:\sqrt{{R}}}\Rightarrow\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\mathrm{1} \\ $$$${S}_{{area}} =\pi\left(\mathrm{1}\right)^{\mathrm{2}} =\pi\:{cm}^{\mathrm{2}} \\ $$…
Question Number 19709 by NECC last updated on 14/Aug/17 $${In}\:{the}\:{cyclic}\:{quadrilateral}\:{ABCD} \\ $$$${AB}=\mathrm{7},{BC}=\mathrm{8},{CD}=\mathrm{8},{DA}=\mathrm{15}. \\ $$$${Calculate}\:{the}\:{angle}\:{ADC}\:{and} \\ $$$${the}\:{length}\:{ofAC}. \\ $$ Answered by Tinkutara last updated on 15/Aug/17…
Question Number 85241 by Power last updated on 20/Mar/20 Answered by jagoll last updated on 20/Mar/20 $$\mathrm{minimum}\:\frac{\mathrm{4s}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }{\mathrm{5st}}\:=\:\frac{\mathrm{4s}^{\mathrm{2}} }{\mathrm{5st}}\:+\:\frac{\mathrm{t}^{\mathrm{2}} }{\mathrm{5st}} \\ $$$$=\:\frac{\mathrm{4}}{\mathrm{5}}\left(\frac{\mathrm{s}}{\mathrm{t}}\right)\:+\:\frac{\mathrm{1}}{\mathrm{5}}\left(\frac{\mathrm{t}}{\mathrm{s}}\right). \\ $$$$\mathrm{let}\:\frac{\mathrm{s}}{\mathrm{t}}\:=\:\mathrm{u}\:\Rightarrow\mathrm{f}\left(\mathrm{u}\right)\:=\:\frac{\mathrm{4}}{\mathrm{5}}\mathrm{u}+\frac{\mathrm{1}}{\mathrm{5}}\mathrm{u}^{−\mathrm{1}}…
Question Number 19699 by Tinkutara last updated on 14/Aug/17 $$\mathrm{Let}\:{S}\:\mathrm{be}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{centre}\:{O}.\:\mathrm{A}\:\mathrm{chord} \\ $$$${AB},\:\mathrm{not}\:\mathrm{a}\:\mathrm{diameter},\:\mathrm{divides}\:{S}\:\mathrm{into}\:\mathrm{two} \\ $$$$\mathrm{regions}\:{R}_{\mathrm{1}} \:\mathrm{and}\:{R}_{\mathrm{2}} \:\mathrm{such}\:\mathrm{that}\:{O}\:\mathrm{belongs} \\ $$$$\mathrm{to}\:{R}_{\mathrm{2}} .\:\mathrm{Let}\:{S}_{\mathrm{1}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{centre}\:\mathrm{in} \\ $$$${R}_{\mathrm{1}} ,\:\mathrm{touching}\:{AB}\:\mathrm{at}\:{X}\:\mathrm{and}\:{S}\:\mathrm{internally}. \\ $$$$\mathrm{Let}\:{S}_{\mathrm{2}}…
Question Number 19668 by Joel577 last updated on 14/Aug/17 Commented by Joel577 last updated on 15/Aug/17 $$\mathrm{An}\:\mathrm{equateral}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{side}\:\mathrm{lenght} \\ $$$${d}.\:\mathrm{Circle}\:{L}_{\mathrm{1}} \:\mathrm{touching}\:\Delta{ABC}\:\mathrm{at}\:{A}\:\mathrm{and}\:{B}. \\ $$$$\mathrm{Circle}\:{L}_{\mathrm{2}} \:\mathrm{touching}\:{AC},\:{BC},\:\mathrm{and}\:{L}_{\mathrm{1}} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{radius}\:\mathrm{of}\:\mathrm{circle}\:{L}_{\mathrm{2}}…