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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 818.
To Anonymous, problem 818: solution is not complete.
alpha=pi/12=15 G x=pi/2+alpha= 105 GI solve system of 3 equationsc=tg(a)c+1=b*tg(3a)c+1=(b+1)*tg(2a)
In the triangle AFE AFE) = x = 180 - alpha -(90 - 2 * alpha) = 90 + alpha.Then (DCE) = (AFE) = 90 + alpha.In the triangle ADE : AD/sin(alpha) = DE/sin(2 alpha) Therefore: DE = 2 AD cos(alpha) ----- (1)In the triangle DCE : DC/sin(2 alpha) = DE/sin(90 + alpha) Therefore: DE = DC / ( 2 sen(alpha) ) ----- (2).DC = ADTherefore from (1) and (2) : 2 cos(alpha) = 1/( 2 sen(alpha) ) ------ (3)If we simplify the (3): sin(2 alpha) = 1/2.Therefore: alpha = 15 x = 90 + 15 = 105
Problem 818: Try to use elementary geometry (Euclid's Elements).
Geometric solution by Michael TsourakakisEC sectionAB=H(q) is Circumcircle of the scare ABCD and (q) section AE=K bisector of the angle EAD section (q) =M.angleDCK=angleKAD=2α .So angleDCM=angleDKM=angleCDK=angleKDM=α.Therefore, angleCBK=αBut angleKBD=angleKAD=2α .So angleDBC=45=2α+α=3α. So,α=15angleCAK=α=angleCEA.So, angleHKA=2α =30From the triangle HCA : x=180-30-45=1050
Daca notam cu P intersectia dreptelor CF si AD vom obtine:2α=m(m( in triunghiul AFP dreptunghic in A, m(x=m(<AFE)=180-(90-α)=90+α
To: Michael Tsourakakis---In your solution you have implied that the ARC CKMD is split into 3 equal parts. This is not true based on your reasoning.
correct solution(q) is Circumcircle of the square ABCD and (q) section AE=KBD ,is mediator of AC and V is mediator of CE. BD section (v)=M.Then , CM is mediator of AE ,so CM is perpendicular of the AE.Beacause AC is ,diameter the circle (q) ,CK ,is perpendicular of AE and the points C,K,M , are collinear . The triangle ACE ,is isosceles.So, angle CAE=α and angleCAD=3α=45 . Τhen,α=15x=180-45-2α =135-30=105 degreeshttp://img191.imageshack.us/img191/5597/geogebra3.png(Michael Tsourakakis )
correction:The triangle CAE is isosceles And not "the triangle ΜΑΕ is isosceles"(Michael Tsourakakis
Anonymous If we say the cut of BD with CF = P and the cut AE with CD =R you have accept without verification that angle CPR= 2 alpha or equivalent that angle APR =90 degree . If you can not verify one of these two saying than the solution is wrong.
you're right.Thanks for the correction.I'll look againHAPPY NEW YEARMichael Tsourakakis
http://img839.imageshack.us/img839/3518/problem818.pngDraw circumcircle of triangle CDENote that angleDOG= 2 α and AE cut arc CD at midpoint GSince angle (GAD)= angle (DOG)= 2 α and AD//GO => ADOG is a parallelogramAnd OD=AD=CD => triangle DOC is a equilateral So 2 α= 30In triangle AFE , we have x=180- α-(90-2 α)= 90+ α= 105 degrees