## Friday, May 16, 2014

### Geometry Problem 1016. Triangle, 45 Degrees, Area, Quadrilateral, Altitude, Bisector

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to enlarge.

1. Let Area(XYZ) denote area of triangle XYZ
Draw altitudes AE’ and CF’ of triangle ABC
Let N is orthocenter of triangle ABC
Triangles ABE’ and CBF’ are right isosceles triangles
Quadrilateral HNE’C is cyclic => ∠ (NHE)= ∠ (NCE)=45
So E’ coincide to E
Similarly F’ coincide to F
Let BM and DL are altitudes of triangles BEF and DEF
Triangle EBF similar to ABC … ( case SAS)
So AB/BE=BC/BF = sqrt(2)= (BM+DL)/BM => DL/BM= sqrt(2)-1
We have S= Area(ABC)= 2.Area(BEF)…. (1)
And Area(DEF)/Area(BEF)=sqrt(2)-1.. ( both triangles have the same base)
So S1=Area(BEF)+Area(DEF)= sqrt(2). Area(BEF)…(2)
From (1) and (2) we have S/S1= sqrt(2) => S^2/S1^2= 2

2. Denote AB+BC=s and AB*BC=p. 2*[FBED]=2*[ABC]-2*[AFD+CED]=p/sqrt(2)-(AF+EC)h, where h is equal height from D to AF and EC. Note that BFHC and BEHA are concyclic, so E and F are feet of altitudes of AE=AB/sqrt(2) and CF=BC/sqrt(2) respectively. Therefore (AF+EC)=s-s/sqrt(2)=s(sqrt(2)-1)/sqrt(2), and h=p/(sqrt(2)*s). Plugging these into first equation yields
2*[FBED]=p/sqrt(2)-p(sqrt(2)-1)/2=sqrt(2)p/2-(sqrt(2)-1)p/2=p/2. 2*[ABC]=p/sqrt(2)
so [ABC]/[FBED]=sqrt(2)

3. Problem 1016
Let S2 = S(ABD) and let S3 = S(CBD)

< CHE = 45, so AHEB is concyclic.
Hence < AEB = < AHB = 90
and so BE = c/√2
Similarly BF = a/√2

Now AD = bc/(a+c) so S2/S = {bc/(a+c)}/b,
Hence S2 = S{c/(a+c)}
Similarly S3 = S{a/(a+c)}

So S1 = S(BFD) + S(BDE) = {(a/ √2)/c}S2 + { (c/√2)/a}S3
S1 = S{(a/√2/(a+c)} + S{ c/√2/(a+c)} = 1/√2
Therefore S^2 /S1^2 = 2.

Sumith Peiris
Moratuwa
Sri Lanka