Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 818.
Wednesday, October 24, 2012
Problem 818: Square, Triangle, Angles
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http://www.wolframalpha.com/input/?i=tg%283a%29%3Dtg%28a%29%2B1
ReplyDeletealpha=0.30714520
x=pi/2+alpha
To Anonymous, problem 818: solution is not complete.
Deletealpha=pi/12=15 G
ReplyDeletex=pi/2+alpha= 105 G
I solve system of 3 equations
c=tg(a)
c+1=b*tg(3a)
c+1=(b+1)*tg(2a)
In the triangle AFE
ReplyDeleteAFE) = 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 = AD
Therefore 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).
ReplyDeleteGeometric solution by Michael Tsourakakis
DeleteEC 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=α
But angleKBD=angleKAD=2α .So angleDBC=45=2α+α=3α. So,α=15
angleCAK=α=angleCEA.So, angleHKA=2α =30
From the triangle HCA : x=180-30-45=1050
http://1.bp.blogspot.com/-H0dtaRbb9bY/UMTFm5yQRyI/AAAAAAAAAAM/5Dco1bojssk/s1600/Untitled.jpg
DeleteDaca notam cu P intersectia dreptelor CF si AD vom obtine:2α=m(m( in triunghiul AFP dreptunghic in A, m(
ReplyDeletex=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.
ReplyDeletecorrect solution
Delete(q) is Circumcircle of the square ABCD and (q) section AE=K
BD ,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=15
x=180-45-2α =135-30=105 degrees
http://img191.imageshack.us/img191/5597/geogebra3.png
(Michael Tsourakakis )
correction:
DeleteThe triangle CAE is isosceles And not "the triangle ΜΑΕ is isosceles"
(Michael Tsourakakis
Anonymous
ReplyDeleteIf 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 again
DeleteHAPPY NEW YEAR
Michael Tsourakakis
http://img839.imageshack.us/img839/3518/problem818.png
ReplyDeleteDraw circumcircle of triangle CDE
Note that angleDOG= 2 α and AE cut arc CD at midpoint G
Since angle (GAD)= angle (DOG)= 2 α and AD//GO => ADOG is a parallelogram
And OD=AD=CD => triangle DOC is a equilateral
So 2 α= 30
In triangle AFE , we have x=180- α-(90-2 α)= 90+ α= 105 degrees
Excellent work Peter
Deletehttp://www.mathematica.gr/forum/viewtopic.php?f=22&t=32386&p=149846
ReplyDeleteDrop a perpendicular from D to AG to meet AG at X and CF at Y. Let AG meet CD at Z.
ReplyDeleteDX is an altitude of right ∆ ADZ, hence < ZDY = DAZ = 2α.
But < DYC = 2α considering ∆GXY, so < DCY also = 90-α.
Therefore DY = DC
So DX = ½ DY = ½ DC = ½ AD
Hence ADX is a 30-60-90 ∆ and 2α = 30 and α = 15
Finally from ∆ AGF, x = 180 - (90 - 2α) - α =90 + α = 105
Sumith Peiris
Moratuwa
Sri Lanka