Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Problem submitted by Charles T.
Click the figure below to see the complete problem 859.
Friday, February 15, 2013
Problem 859: Isosceles Right Triangle, Angles,18, 45 Degrees, Congruence, Perpendicular bisector
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Answer: 27
ReplyDeleteSorry for using trigonometry but i can't find a 'normal' solution.
Let AB=BC=DE=AD=DF=2
Draw a height BG from B to AC. It is clear from the figure that <DBG=126, <BGE=135 and BG=root(2). From sine theorem BD=4*sin9 and DE=2.
From cosine theorem(I leave to the reader to use this theorem in triangles DBG and DEG)we see that DEBG is a inscribed quadrilateral.
So <EDB=45 and <DEF=54. So x=180-2*54-45=27 degrees.
Let is ,AD=DE=DF=AB=BC=α and c , the circle (A,α).
ReplyDeleteThe line DE , meets the circle c , in point H and the line AE meets the HC in point P. Then , AH = α
The line EF (perpendicular bisector of BC) meets AC in point M , midpoint of AC and E is midpoint of AP.
So , frome the triangle ACP we have ,EM//PC . Therefore , HC //AB . So the quadrilateral ABCH we have : ΗΑ=ΑΒ=ΒC=α , HC , AB , perpendicular to BC , so, ABCH is a square.( The proof is easy)
From the isosceles triangle DAH , angle DAH = 90+18=108 degrees , so, angle DAH=36 degrees. But , from the isosceles triangle ADB , angle ADB=81 degrees ,therefore, angle LDB=81-36=45 degrees.
From the triangle ADL , angle ALE =18+36= 54 degrees . Because AL // EF, we have , angle LEF=54 degrees .So , from the isosceles triangle DEF , we have, 45+X+54+54=180 , therefore
X=27 degrees .
Plan : http://img825.imageshack.us/img825/6397/geogebra5.png
Michael
DeleteIt is not clear how EM//PC from line 3 and 4 of your solution :"The line EF (perpendicular bisector of BC) meets AC in point M , midpoint of AC and E is midpoint of AP"
Please provide more details.
Peter Tran
Hello Peter.You are right.
Deletehasty my solution.
I give you one of the two solutions that I've done
Let is ABCB’ a square. Then , FE is perpendicular bisector of AB’. So AE=EB’=x.
If, AD=AB=DE=DF=R and c=(A,R) , c’=(D,R) two circles ,then for circle c’ we have
R^2-AE^2=EB’.DE , R^2-x^2=x.R .Therefore, x=R[(sqrt(5)-1]/2
So ,for circle c’ , with radius R, the chord AE=x length is R[(sqrt(5)-1]/2 .But this is the chord length of a regular decagon .So , the central angle ADE=36 degrees.
Then , angle ALE =18+36=54 degrees .But AB//FE ,so, angle LEF = angle DFE =54 degrees.
Additional ,for isosceles triangle ADB, is, angle ADB=81,therefore, angle LDB=45 degrees.
Finality ,for triangle EDF 54+54+45+X=180 , so, x=27 degrees
I hope I have not accidentally
plan. http://img407.imageshack.us/img407/2303/geogebra8.png
The correct address for the image
ReplyDeletehttp://img145.imageshack.us/img145/6397/geogebra5.png
{P}= cercul (D,AD) taie pe AC=> <EDA=y
ReplyDelete{ T}= BA cu DE,<BTD=18+x
BA||EF, <BTD=<DEF <EDF=144-2y
x==144-2y-(81-y)=63-y unde 0<y<54
Is y always = 36 or not?
DeleteI extended DB until it intersects EF. I call that point G. I note that DG=DB + BG. I determine that <DGF=99deg. I note that sin81deg = sin99deg. I use the law of sines and substitution to conclude that <F=54 deg. <X = 180 - 99 - 54 = 27 deg.
ReplyDeleteI can scan my solution with drawing to anyone who is interested. All that is required is to extend line DB until it intersects with line EF to form 2 other triangles. Then use the law of sines and algebraic substitution.
ReplyDeleteLet AB = BC = AD = DE = DF = a
ReplyDeleteDraw DMN perpendicular to AB (or EF)
meeting AB at M and EF at N.
Note <ABD = <ADB = 81 deg.
So <MDB = 90 – 81 = 9 deg.
From the isosceles ΔABD,
we have BD = 2a sin 9
Follows DM = BD cos 9
= 2a sin 9 cos 9 = a sin 18.
EF bisects BC and is pependicular to it.
So MN = a/2.
So DN = DM + MN = a (sin 18 + 1/2)
= a[(sqrt(5) – 1)/4) + 1/2]
= a (sqrt(5) + 1)/4 = a cos 36.
DN bisects <EDF. So <NDF = x + 9
and DN = a cos (x + 9).
Hence a cos (x + 9) = a cos 36,
cos (x + 9) = cos 36,
x + 9 = 36,
x = 27 degrees
aceasta e solutia corecta ,ceruta
ReplyDelete{P}= cercul (D,AD) taie pe AC=> <EDA=y
{ T}= BA cu DE,<BTD=18+y
BA||EF, <BTD=<DEF <EDF=144-2y
x==144-2y-(81-y)=63-y unde 0<y<54
Reply
To Prof Radu Please review your solution. In this problem we don't need to know "y" to determine "x". Thanks.
DeleteValorile lui x sunt in functie de valorile lui y,unde 0x=27
Deletedaca y=30 =>x=33 etc. In problema punctul E nu e precizat el se poate deplaseasa pe arcul de cerc AP
{P}= cercul (D,AD) taie pe AC al acestui cerc.Va rog sa verificati.
Daca y=36,atunci x=27 Daca y=30 =>x=33 etc.
ReplyDeleteunde 0<y<54,<EDA=y.
In Figure http://img145.imageshack.us/img145/6397/geogebra5.png.
ReplyDeletesince <BHD=9 and <EMH=45, then <DLB=<DEM=54 and <ADL=36. Hence <LDB=45.
Since DE=DF, and so <DEF=<DFE=54, <EDF=72.
Furthermore, since <BDL=45, <LDF=72, we see that x=27.
Michael Tsourakakis and 배덕락: Why are D, E, B1, collinear?
ReplyDeleteIf you cannot prove that it is true, the problem is not solved as you claim.
Erina-NJ
http://www.mathematica.gr/forum/viewtopic.php?f=22&t=35375&p=162991
ReplyDeleteLet EF,BP,AE respectively meet AC,DE,QC at P,Q,R.
ReplyDeleteEF is the perpendicular bisector of BC, hence AP = PC so AE = ER which
implies that EF // QC.
Hence < QCR = 45 and so PQ = PC = PB = PA and so ABCQ is a square.
Therefore < DAQ = 108, < AQD = ADQ = 36 and so < DEN = 18+36 = 54 = < DFE.
Hence < EDF = 72 and finally x = 72 – (81-36) = 27
Sumith Peiris
Moratuwa
Sri Lanka