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Category: Geometry

Consider-the-quadrilateral-ABCD-The-points-M-N-P-and-Q-are-the-midpoints-of-the-sides-AB-BC-CD-and-DA-Let-X-AP-BQ-Y-BQ-CM-Q-CM-DN-and-T-DN-AP-Prove-that-XYZT-AQX-BMY-

Question Number 16946 by Tinkutara last updated on 28/Jun/17 $$\mathrm{Consider}\:\mathrm{the}\:\mathrm{quadrilateral}\:{ABCD}. \\ $$$$\mathrm{The}\:\mathrm{points}\:{M},\:{N},\:{P}\:\mathrm{and}\:{Q}\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:{AB},\:{BC},\:{CD} \\ $$$$\mathrm{and}\:{DA}. \\ $$$$\mathrm{Let}\:{X}\:=\:{AP}\:\cap\:{BQ},\:{Y}\:=\:{BQ}\:\cap\:{CM}, \\ $$$${Q}\:=\:{CM}\:\cap\:{DN}\:\mathrm{and}\:{T}=\:{DN}\:\cap\:{AP}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\left[{XYZT}\right]\:=\:\left[{AQX}\right]\:+\:\left[{BMY}\right] \\ $$$$+\:\left[{CNZ}\right]\:+\:\left[{DPT}\right]. \\…

Six-points-A-B-C-D-E-and-F-are-placed-on-a-square-rigid-as-shown-How-many-triangles-that-are-not-right-angled-can-be-drawn-by-using-3-of-these-6-points-as-vertices-

Question Number 16940 by Tinkutara last updated on 28/Jun/17 $$\mathrm{Six}\:\mathrm{points}\:\mathrm{A},\:\mathrm{B},\:\mathrm{C},\:\mathrm{D},\:\mathrm{E},\:\mathrm{and}\:\mathrm{F}\:\mathrm{are} \\ $$$$\mathrm{placed}\:\mathrm{on}\:\mathrm{a}\:\mathrm{square}\:\mathrm{rigid},\:\mathrm{as}\:\mathrm{shown}. \\ $$$$\mathrm{How}\:\mathrm{many}\:\mathrm{triangles}\:\mathrm{that}\:\mathrm{are}\:\boldsymbol{\mathrm{not}} \\ $$$$\mathrm{right}-\mathrm{angled}\:\mathrm{can}\:\mathrm{be}\:\mathrm{drawn}\:\mathrm{by}\:\mathrm{using}\:\mathrm{3} \\ $$$$\mathrm{of}\:\mathrm{these}\:\mathrm{6}\:\mathrm{points}\:\mathrm{as}\:\mathrm{vertices}? \\ $$ Commented by Tinkutara last updated…

Please-solve-Q-16066-or-please-tell-me-whether-to-post-its-solution-or-not-I-don-t-understood-the-solution-

Question Number 16911 by Tinkutara last updated on 28/Jun/17 $$\mathrm{Please}\:\mathrm{solve}\:\mathrm{Q}.\:\mathrm{16066}\:\mathrm{or}\:\mathrm{please}\:\mathrm{tell}\:\mathrm{me} \\ $$$$\mathrm{whether}\:\mathrm{to}\:\mathrm{post}\:\mathrm{its}\:\mathrm{solution}\:\mathrm{or}\:\mathrm{not}.\:\mathrm{I} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{understood}\:\mathrm{the}\:\mathrm{solution}. \\ $$ Commented by mrW1 last updated on 28/Jun/17 $$\mathrm{I}\:\mathrm{think}\:\mathrm{I}\:\mathrm{have}\:\mathrm{the}\:\mathrm{answer}.\:\mathrm{I}'\mathrm{ll}\:\mathrm{post} \\…

In-plane-parallel-lines-are-the-lines-which-don-t-meet-each-other-What-is-the-condition-in-space-that-two-lines-be-parallel-

Question Number 16882 by RasheedSoomro last updated on 27/Jun/17 $$\:^{\bullet} \mathrm{In}\:\mathrm{plane}\:\mathrm{parallel}\:\mathrm{lines}\:\mathrm{are}\:\mathrm{the}\:\mathrm{lines} \\ $$$$\mathrm{which}\:\mathrm{don}'\mathrm{t}\:\mathrm{meet}\:\mathrm{each}\:\mathrm{other}. \\ $$$$ \\ $$$$\:^{\bullet} \mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{in}\:\mathrm{space} \\ $$$$\mathrm{that}\:\mathrm{two}\:\mathrm{lines}\:\mathrm{be}\:\mathrm{parallel}? \\ $$ Commented by prakash…

Let-ABCD-be-a-parallelogram-The-points-M-N-and-P-are-chosen-on-the-segments-BD-BC-and-CD-respectively-so-that-CNMP-is-a-parallelogram-Let-E-AN-BD-and-F-AP-BD-Prove-that-AEF-DFP-

Question Number 16879 by Tinkutara last updated on 27/Jun/17 $$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{parallelogram}.\:\mathrm{The} \\ $$$$\mathrm{points}\:{M},\:{N}\:\mathrm{and}\:{P}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{segments}\:{BD},\:{BC}\:\mathrm{and}\:{CD}, \\ $$$$\mathrm{respectively},\:\mathrm{so}\:\mathrm{that}\:{CNMP}\:\mathrm{is}\:\mathrm{a} \\ $$$$\mathrm{parallelogram}.\:\mathrm{Let}\:{E}\:=\:{AN}\:\cap\:{BD}\:\mathrm{and} \\ $$$${F}\:=\:{AP}\:\cap\:{BD}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\left[{AEF}\right]\:=\:\left[{DFP}\right]\:+\:\left[{BEN}\right]. \\ $$ Terms…

From-a-point-on-the-circumcircle-of-an-equilateral-triangle-ABC-parallels-to-the-sides-BC-CA-and-AB-are-drawn-intersecting-the-sides-CA-AB-and-BC-at-the-points-M-N-P-respectively-Prove-that-the

Question Number 16877 by Tinkutara last updated on 27/Jun/17 $$\mathrm{From}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of}\:\mathrm{an} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}\:\mathrm{parallels}\:\mathrm{to} \\ $$$$\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB}\:\mathrm{are}\:\mathrm{drawn}, \\ $$$$\mathrm{intersecting}\:\mathrm{the}\:\mathrm{sides}\:{CA},\:{AB}\:\mathrm{and}\:{BC} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{points}\:{M},\:{N},\:{P},\:\mathrm{respectively}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{points}\:{M},\:{N}\:\mathrm{and}\:{P}\:\mathrm{are} \\ $$$$\mathrm{collinear}. \\ $$ Terms…

Let-P-be-a-point-on-the-circumcircle-of-the-equilateral-triangle-ABC-Prove-that-the-projections-of-any-point-Q-onto-the-lines-PA-PB-and-PC-are-the-vertices-of-an-equilateral-triangle-

Question Number 16878 by Tinkutara last updated on 27/Jun/17 $$\mathrm{Let}\:{P}\:\mathrm{be}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circumcircle}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{equilateral}\:\mathrm{triangle}\:{ABC}.\:\mathrm{Prove} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{projections}\:\mathrm{of}\:\mathrm{any}\:\mathrm{point}\:{Q} \\ $$$$\mathrm{onto}\:\mathrm{the}\:\mathrm{lines}\:{PA},\:{PB}\:\mathrm{and}\:{PC}\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{vertices}\:\mathrm{of}\:\mathrm{an}\:\mathrm{equilateral}\:\mathrm{triangle}. \\ $$ Commented by prakash jain last…

Let-P-1-P-2-P-n-be-a-convex-polygon-with-the-following-property-for-any-two-vertices-P-i-and-P-j-there-exists-a-vertex-P-k-such-that-the-segment-P-i-P-j-is-seen-from-P-k-under-an-ang

Question Number 16875 by Tinkutara last updated on 27/Jun/17 $$\mathrm{Let}\:{P}_{\mathrm{1}} ,\:{P}_{\mathrm{2}} ,\:…,\:{P}_{{n}} \:\mathrm{be}\:\mathrm{a}\:\mathrm{convex}\:\mathrm{polygon} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{following}\:\mathrm{property}\::\:\mathrm{for}\:\mathrm{any} \\ $$$$\mathrm{two}\:\mathrm{vertices}\:{P}_{{i}} \:\mathrm{and}\:{P}_{{j}} ,\:\mathrm{there}\:\mathrm{exists}\:\mathrm{a} \\ $$$$\mathrm{vertex}\:{P}_{{k}} \:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{segment}\:{P}_{{i}} {P}_{{j}} \\ $$$$\mathrm{is}\:\mathrm{seen}\:\mathrm{from}\:{P}_{{k}}…

Let-I-be-the-incenter-of-ABC-It-is-known-that-for-every-point-M-AB-one-can-find-the-points-N-BC-and-P-AC-such-that-I-is-the-centroid-of-MNP-Prove-that-ABC-is-an-equilateral-triangle-

Question Number 16873 by Tinkutara last updated on 27/Jun/17 $$\mathrm{Let}\:{I}\:\mathrm{be}\:\mathrm{the}\:\mathrm{incenter}\:\mathrm{of}\:\Delta{ABC}.\:\mathrm{It}\:\mathrm{is} \\ $$$$\mathrm{known}\:\mathrm{that}\:\mathrm{for}\:\mathrm{every}\:\mathrm{point}\:{M}\:\in\:\left({AB}\right), \\ $$$$\mathrm{one}\:\mathrm{can}\:\mathrm{find}\:\mathrm{the}\:\mathrm{points}\:{N}\:\in\:\left({BC}\right)\:\mathrm{and} \\ $$$${P}\:\in\:\left({AC}\right)\:\mathrm{such}\:\mathrm{that}\:{I}\:\mathrm{is}\:\mathrm{the}\:\mathrm{centroid}\:\mathrm{of} \\ $$$$\Delta{MNP}.\:\mathrm{Prove}\:\mathrm{that}\:{ABC}\:\mathrm{is}\:\mathrm{an} \\ $$$$\mathrm{equilateral}\:\mathrm{triangle}. \\ $$ Terms of Service…