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Category: Geometry

Let-AC-be-a-line-segment-in-the-plane-and-B-a-point-between-A-and-C-Construct-isosceles-triangles-PAB-and-QBC-on-one-side-of-the-segment-AC-such-that-APB-BQC-120-and-an-isosceles-triangle-RAC-

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} \\…

Let-ABCD-be-a-parallelogram-Two-points-E-and-F-are-chosen-on-the-sides-BC-and-CD-respectively-such-that-EB-EC-m-and-FC-FD-n-Lines-AE-and-BF-intersect-at-G-Prove-that-the-ratio-

Question Number 19238 by Tinkutara last updated on 07/Aug/17 $$\mathrm{Let}\:{ABCD}\:\mathrm{be}\:\mathrm{a}\:\mathrm{parallelogram}.\:\mathrm{Two} \\ $$$$\mathrm{points}\:{E}\:\mathrm{and}\:{F}\:\mathrm{are}\:\mathrm{chosen}\:\mathrm{on}\:\mathrm{the}\:\mathrm{sides} \\ $$$${BC}\:\mathrm{and}\:{CD},\:\mathrm{respectively},\:\mathrm{such}\:\mathrm{that} \\ $$$$\frac{{EB}}{{EC}}\:=\:{m},\:\mathrm{and}\:\frac{{FC}}{{FD}}\:=\:{n}.\:\mathrm{Lines}\:{AE}\:\mathrm{and}\:{BF} \\ $$$$\mathrm{intersect}\:\mathrm{at}\:{G}.\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{ratio} \\ $$$$\frac{{AG}}{{GE}}\:=\:\frac{\left({m}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{1}\right)}{{mn}}. \\ $$ Commented by ajfour…

Question-84718

Question Number 84718 by Power last updated on 15/Mar/20 Answered by mr W last updated on 15/Mar/20 $${AC}^{\mathrm{2}} ={PA}^{\mathrm{2}} +{PC}^{\mathrm{2}} −\mathrm{2}×{PA}×{PC}\:\mathrm{cos}\:\mathrm{120}° \\ $$$$\Rightarrow{AC}^{\mathrm{2}} ={PA}^{\mathrm{2}} +{PC}^{\mathrm{2}}…

A-semicircle-is-tangent-to-both-legs-of-a-right-triangle-and-has-its-centre-on-the-hypotenuse-The-hypotenuse-is-partitioned-into-4-segments-with-lengths-3-12-12-and-x-as-shown-in-the-figure-Det

Question Number 19150 by Tinkutara last updated on 06/Aug/17 $$\mathrm{A}\:\mathrm{semicircle}\:\mathrm{is}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{both}\:\mathrm{legs}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{right}\:\mathrm{triangle}\:\mathrm{and}\:\mathrm{has}\:\mathrm{its}\:\mathrm{centre}\:\mathrm{on}\:\mathrm{the} \\ $$$$\mathrm{hypotenuse}.\:\mathrm{The}\:\mathrm{hypotenuse}\:\mathrm{is} \\ $$$$\mathrm{partitioned}\:\mathrm{into}\:\mathrm{4}\:\mathrm{segments},\:\mathrm{with}\:\mathrm{lengths} \\ $$$$\mathrm{3},\:\mathrm{12},\:\mathrm{12},\:\mathrm{and}\:{x},\:\mathrm{as}\:\mathrm{shown}\:\mathrm{in}\:\mathrm{the}\:\mathrm{figure}. \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:'{x}'. \\ $$ Commented by Tinkutara…

Question-84689

Question Number 84689 by mr W last updated on 15/Mar/20 Commented by mr W last updated on 15/Mar/20 $${the}\:{parabola}\:{is}\:{rolled}\:{along}\:{the}\:{circle} \\ $$$${as}\:{shown}. \\ $$$${find}\:{the}\:{equation}\:{of}\:{the}\:{parabola}\:{when} \\ $$$${the}\:{touching}\:{point}\:{is}\:{at}\:{the}\:{position}\:\theta.…

Question-84655

Question Number 84655 by ajfour last updated on 14/Mar/20 Commented by ajfour last updated on 14/Mar/20 $${If}\:{area}\:{of}\:\bigtriangleup{ABC}\:\:{is}\:{equal}\:{to} \\ $$$${max}\:{trapezium}\:{area}\:{BCED}\:, \\ $$$${find}\:{r}\:{in}\:{terms}\:{of}\:{ellipse}\: \\ $$$${parameters}\:{a}\:{and}\:{b}. \\ $$…