Saturday, October 17, 2009

Problem 367. Square, Angle, 45 degrees, Pythagoras, Congruence, Similarity

Proposed Problem
Click the figure below to see the complete problem 367 about Square, Angle, 45 degrees, Pythagoras, Congruence, Similarity.

See more:
Complete Geometry Problem 367
Level: High School, SAT Prep, College geometry

8 comments:

1. AB=BC=CD=AD=(a+b+x)/sqrt(2)...(*)

from similarity between triangles AGD and HGA:
AG=sqrt(x^2+xb)...(**)

from similarity between triangles ABH and GAH:

AB/GA=AH/GH=BH/AH

by setting the values from (*) and (**); we get:
AH=x*[(a+b+x)/sqrt(2)]/sqrt(x^2+bx)...(1).
and
AH=sqrt(x^2+xa)...(2).

after solving the equations (1),(2) we get:

x^2=a^2+b^2

2. http://geometri-problemleri.blogspot.com/2009/11/problem-47-ve-cozumu.html

3. since angle(GAF)=angle(GDF)=45 and angle(ADF)=90, GADF is a cyclic and so EGHF is an also cyclic. Hence two triangles BEG and BHC are similar and hence a(a+x)=BE(a+b+x)/sqrt(2).
Since BE=(a+b+x)/{sqrt(2)(x+b)}, we have
get x^2 =a^2 +b^2.

4. There is a very simple and elegant solution if we reflect triangle ABE across AE so B becomes B' and we reflect triangle ADF across AF so that D becomes D', then since AB=AD we must have B'=D'. Also angle GB'H=angle GB'A+ angle HB'A=angleGBA+ angle HDA=45+45+90, so by Pythag on triangle GB'H, a^2+b^2=x^2.

5. Because angle EAH=angle HBE=45, the quadrilateral BAHE is, cyclic quadrilateral.So, The EH is perpendicular to AF.
Because angle GDF=angle GAH=45, the quadrilateral AGFD is, cyclic quadrilateral.So, the FG is perpendicular to AE. Therefore, The EH, FG, is heights of triangle EAF, so, ALK, is the other height of the triangle AEF
By cyclic quadrilaterals arising, we have, angle EKG=angle FKH=45,so, angle GKH=90.
More , angle AEF= angle GHA=angle BEA and angle KFH=angle AGD=angle AFD .
Therefore, triangle BEG=triangle GEK and triangle KHF=triangle FDH.So ,GK=GB=a and KH=HD=b
Pythagoras theorem to the triangle GKH: x^2=a^2+b^2
Image: http://img267.imageshack.us/img267/2807/geogebra.png

6. Tr.GAH & tr.ABH are similar(<GAH=<ABH=45,<H is common).so we have,AG/AB=AH/a+x.......I)
Tr.GAH & tr.ADG are similar(<GAH=<ADG=45,<G is common).so we have AH/AD=AG/b+x......II)
multiplying the eq.s,we get AG*AH/AB*AD=AG*AH/(a+x)(b+x)
therefore,(a+x)(b+x)=AB*AD=c^2[let AB=AD=c]
therefore,2ab+2ax+2bx+2x^2=2c^2=(c*sqrt.2)^2=BD^2=(a+x+b)^2
therefore,2x^2=a^2+x^2+b^2
so we conclude, x^2=a^2+b^2

7. Find M such that DM = a and < ADM = 45.
Tr.s ADM & AEB are congruent from which we deduce that Tr.s AGH & AMH are congruent as well both cases SAS.
Hence FM = x and applying Pythagoras to Tr. HDM the result follows.

Sumith Peiris
Moratuwa
Sri Lanka

8. See the drawing

Define H’ rotation of H with angle –90° and center A
=>AH=AH’ and BH’=DH=b
=> BH’ ⊥DH <=> BH’ ⊥BG
∠HAG=45° and ∠HAH’=90° => ∠GAH’=45°
ΔGAH is congruent to ΔGAH’ (SAS)
=> GH’=GH=x
BH’=b, BG=a, BH’ ⊥BG and GH’=x => ΔGBH’ is a rectangle triangle in B
Therefore x^2=a^2+b^2