Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Friday, July 7, 2017
Geometry Problem 1338: Four Squares, Diagonals, Angle, 45 Degrees, Areas
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https://goo.gl/photos/dB3G7J8HWSRkdPrx9
ReplyDeleteLet u= ∠ (FAD) and v= ∠ (BAE)
Create triangle ADP such that triangle ABE congruent to tri. ADP( see sketch)
We have ∠ (DAP)= v and ∠ (ADP)= 45 and AE=AP, BE= DP
U+v= 45 and FD⊥DP
Triangle AEF congruent to tri. APF ( case SAS)
So EF= FP
Applying Pythagoras theorem in triangle FDP
FP^2= FD^2+DP^2 = EF^2= FD^2+BE^2
From this result we will have S= S1+S2
Problem 1338
ReplyDeleteLet AE ,BC intersects in K (K on BC) and AF,CD intersects in L ( L on CD).
Let AM ( M on extention CD) with AM perpendicular in AE.
Now triangle ABK=triangle ADM.Let BN perpendicular in FD (N on AM).
But triangle ABE=triangle and AE=AN.
Is triangle AEF=triangle AFN. So EF=FN, <FDN =90.
FN^2=FD^2+DN^2 or EF^2/2=FD^2/2+BE^2/2 or S=S1+S2.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE
Find G such that < GAE = < GDF = 90
ReplyDeleteLet BE = a, EF = b and FD = c
< FAG = 45 and < BAE = < DAG so Tr.s ABE & ADG are congruent ASA
So AG = AE
Hence Tr.s AEF & AGF are congruent SAS and hence FG = b
Applying Pythagoras to Tr. DFG,
a^2+c^2 = b^2
But S1 = a^2/4 and similar for b^2 and c^2
So S1+S2 =S
Sumith Peiris
Moratuwa
Sri Lanka
Let the lower left point of the yellow square be M.
ReplyDeleteBy use of the inscribe angle theorem we see that M is the center of the circumscribed circle of triangle AEF. As a result AM = EM.
Let N be the intersection of the AD and the line through EM.
Now apply Pythagoras on triangle ANM
Good straightforward proof
DeleteTop Class Deduction....!!
DeleteExcellent solution. Thank you
DeleteWe shall prove BE^2+DF^2=EF^2 (*), equivalent to relation to be proven.
ReplyDeleteLet G reflection of B in AE; easily we notice it is also the reflection of D in AF, thus by symmetry <AGE=<ABE=45 and GE=BE, <AGF=<ADF=45 and FG=FD hence
<EGF=90. By Pythagorean theorem in triangle EGF we get (*), thus we are done.
Best regards
spin the figure 90 degrees clockwise around A.
ReplyDeleteLet E' be the point correponding to the rotation of E.
E'D = EB cause D is rotation of B.
triangle FE'D is rectangle in D.
Triangle EAF =Triangle AFE' AF is common, AE=AE' and both have 45 degree between these sides
Apply Pythagorean theorem to FE'D
Let the lower left points of S1 and S2 be K and L respectively, the top right point of S be T, and AH be the altitude from A, i.e. AH _|_ BD and H is on BD. Since AK/AH = AE/AF = AH/AL => AH^2 = AK.AL (1). Since AB = AD, the area of triangle ABD = AH.BD/2 = AH^2 (2). Using (1) and (2) we have that area of AKTL = AK.AL = AH^2 = area of triangle ABD = area of AKEFL + area of triangle KBD + area of triangle LFD = (area of AKTL - S/2) + S1/2 + S2/2 => 0 = -S/2 + S1/2 + S2/2 => S = S1 + S2.
ReplyDeleteSee the drawing
ReplyDeleteDefine a,b and x respectively hypothenuse of S1, S2 and S
S1 is a square => S1= a^2/2
In the same way, S2= b^2/2 and S= x^2/2
From Pb 367 we know that x^2=a^2+b^2
=> (x^2)/2=(a^2)/2+(b^2)/2
Therefore S=S1+S2
Denote the lower left corner of yellow square as O.
ReplyDeleteOE=OF and m(EOF)=2.m(EAF) => O is the circumcenter of EAF.
Hence AO=EO
Let P be the foot of perpendicular from O to AB
Hence OP= length of pink square, PA = length of green square and AO=Length of yellow square
=> S=S1+S2