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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Online Geometry theorems, problems, solutions, and related topics.

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