See complete Problem 188 at:
www.gogeometry.com/problem/p188_square_45_degrees_problem.htm
Square, Diagonal, 45 Degrees Angle. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Friday, October 10, 2008
Elearn Geometry Problem 188
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O is the midpoint of GH, EO cuts CD at E', FO cuts BC at F'
ReplyDeleterotate ⊿ADH about A such that AD and AB coincide and H maps to H'
⊿AGH'≡⊿AGH (SAS), ∠BGA＝∠AGH and ∠GHA＝∠AHD
E,A,D,H are concyclic, HE⊥GA, so is GF⊥AH
G,E,F,H are concyclic with center O
OG＝OE, ∠OEG＝∠EGO＝∠BGE, EE'⊥CD, so is FF'⊥BC
a²+b²＝2(CE'²+CF'²)＝2(x/2)² ⇒ x²=2(a²+b²)
Here's another way to do this: Let A,B,C & D be (0,0),(0,s),(s,s) and (s,0). Also let BG =t. We can, therefore, say that: AG is y=sx/t and BD:x+y=s. Hence, E is (st/(s+t),s^2/(s+t)) Thus, BE^2= a^2 = (st/(s+t))^2 + (s-s^2/(s+t))^2=2[st/(s+t)]^2. Now let DH =q. Hence slope AH=q/s but the angle between AG and AH is 45 deg. Hence [s/t -q/s]/[1+sq/ts]= tan(45)=1. Hence slope DH: (s-t)/(s+t). We locate be F and determine DF^2=b^2 - (s-t)^2. Thus 2(a^2+b^2)= (2st/(s+t))^2 +(s-t)^2 = (s^2+t^2)^2/(s+t)^2 ----(1)
ReplyDeleteWhile x^2 =(t-s)^2+(s-s(s-t)/(s+t))^2=(s^2+t^2)^2/(s+t)^2 ----(2).
By (1) & (2) we can say that x^2=2(a^2+b^2)
Ajit: ajitathle@gmail.com
can i do a question ,please
ReplyDeleteis CH=HD ?
ReplyDeleteCH=HD is not an hypothesis.
ReplyDeleteCharxith said...
ReplyDeleteLet point H be on point D and point G on point C, having GH = DC, AG = BD, and angle CAD = angle GAD. This means that x = DC and that b^2 = 0.
This means x^2 = 2a^2, and since a = AC/2, 2a^2 = a^2 + a^2, segments AC and BD intersect at center of square creating 4 isosceles right tri,
we can conclude that a^2 + a^2 = x^2, or 2a^2 = x^2, by pythagorean theorem.
Charxith said...
ReplyDeleteTherefore from my comment before, x^2 = 2(a^2 + b^2), since b^2 = 0.
Join AC and let m(HAD)=x
ReplyDelete=>m(CAH)=45-x, m(GAC)=x and m(BAG)=45-x
Observe that m(BAE)=m(CAH)=45-x, m(ABE)=m(ACH)=45
=> ABE and ACH are similar triangles
=> AB/BE=AC/CH
=> CH=Sqrt(2)a -----------(1)
Similarly the triangle ADF and ACG are similar and GC=Sqrt(2)b--------(2)
Applying Pythagorean to triangle GCH
=> GH^2=CH^2+GC^2
=> GH^2=2(a^2+b^2)
Q.E.D
See graph here.
ReplyDeleteEBC and EBA are SAS, so are FDA and FDC, therefore ∠ECF = ∠EAF = 45°.
∠EFH = 45° + ∠DAF = ∠AGB and ∠FEG = 45° + ∠BAG = ∠AHD => E, G, C, F and H are concyclic of diameter GH (since GCH rectangle in C).
Let I middle of GH the center of EGCHF and J the middle of arc GEFH.
GJ, JH and EF intercepting arcs of 45° are of equal lengths a∧2+b∧2 (from problem 367, easily seen by rotating ABE clockwise by 90° to form quadrilateral AFDE’ in which FDE’ is rectangle in D, FD = b, DE’ = a and FE’ = EF since EAF and FAE’ are SAS).
So in rectangular triangle GJH, x∧2 = 2(a∧2+b∧2).
The sentence "∠EFH = 45° + ∠DAF = ∠AGB and ∠FEG = 45° + ∠BAG = ∠AHD => E, G, C, F and H are concyclic of diameter GH (since GCH rectangle in C) " is incorrect and needs to be amended as follows:
DeleteC is symmetric to A with respect to BD ⇒
- ∠EFC = ∠EFA = 45°+ ∠DAF = ∠AGB = π - ∠EGC ⇒ E, C ,F, G concyclic and
- ∠ECH = ∠ECF + ∠FCD = 45° + ∠FAD = ∠DFH = π - ∠EFH ⇒ E, C ,F, H concyclic.
So E, F,C, G, H are concyclic of diameter GH (since GCH is right in C).
The rest of the proof unchanged.
And the diagram is corrected as follows.
Let l be the length of the square
ReplyDeleteIt can be derived that EF=Sqrt(a^2+b^2)
=>l=(a+b+Sqrt(a^2+b^2))/Sqrt(2) -----(1)
Let AE=u and AF=v
since m(GAF)=m(GBF)=> ABGF is concyclic and GF=AF=v
Similarly AEHD is concyclic and EH=AE=u
Apply Ptolmey's to ABGF
BG.v+AB.v=BF.Sqrt(2).v
=>BG=Sqrt(2).BF-l
=>BG=Sqrt(2)(a+Sqrt(a^2+b^2))-l -------(2)
Similarly DH=Sqrt(2)(b+Sqrt(a^2+b^2))-l -------(3)
It can be easily derived that x=BG+DH (EH,GF are altitudes. The triangle formed by altitude from A and AG is reflection of ABG along AG. Similarly ADH along AH )
=>x=Sqrt(2)(a+b)+2Sqrt(2)Sqrt(a^2+b^2)-2l (substitute l from (1))
=>x=Sqrt(2)Sqrt(a^2+b^2)
=>x^2=2(a^2+b^2)
See the drawing
ReplyDeleteAD⊥AH => A, D, H are concyclic
Arc EH : ∠EDH=45° and ∠EAH=45° => A, D, H and E are concyclic
Arc AE : ∠EDA=45° => ∠EHA=45° => EA⊥EH and EA=EH
AB⊥BG => A, B, G are concyclic
Arc GF : ∠GBF=45° and ∠GAF=45° => A, B, G and F are concyclic
Arc AG : ∠ABG=90° => ∠AFG=90° => FA⊥FG and FA=FG and AGF=45°
CG⊥CH => C, G, H are concyclic with center O
Arc GH : ∠GCH=90° and ∠GFH=90° => C, G, H and F are concyclic
Arc EF : ∠EHF=45° and ∠EGF=45° => A, B, G, F and E are concyclic and ∠ECF=45°
∠ECF=45° => ∠EOF=90°
A, B, G, F and E are concyclic => OG=OE=OF=OH=OC
x=2OG=2OE
EF^2=OE^2+OF^2=2OE^2=x^2/2
From Pb 367 : EF^2=a^2+b^2
=> a^2+b^2= x^2/2
Therefore x^2=2(a^2+b^2)
Sorry rv but A, B, G, F and E are not concyclic
ReplyDeleteYou are right Greg!
ReplyDeleteI wanted to say "C, G, H, F and E are concyclic" instead of "A, B, G, F and E are concyclic"
Sorry for this mistake
Here’s my solution…
ReplyDeletehttps://youtu.be/WEwKft7Zjkc