Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Details: Click on the figure below.

## Friday, July 7, 2017

### Geometry Problem 1338: Four Squares, Diagonals, Angle, 45 Degrees, Areas

Labels:
45 degrees,
area,
diagonal,
square

Subscribe to:
Post Comments (Atom)

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

DeleteVery nice!

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

ReplyDeletedrawingDefine 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+S2Denote 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