## Saturday, May 1, 2021

### Geometry Problem 1493: Four Squares, Parallelogram, Auxiliary Lines

Geometry Problem 1493. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below. 1. https://photos.app.goo.gl/6SQsFetQuy1gT2zc6

Draw points P and Q per attached sketch
Right triangle AGF congruence to LPA ( case HA)
So PA=FG= v and PL=AG=u+v
Triangle AJB congruence to QFL ( case SAS)
So LF=BJ and ^(QFL)=^(AJB)
So BLFJ is a parallelogram

2. Solution using Pythagoras and Trigonometry

Let AD = a, DG = GJ = b, AF = c, BJ = x, FL = y

Now c^2 = a^2 + 2ab + 2b^2 ....(1) ( Rt.Tr. AFG)
x^2 = 2a^2 + 4ab + 4b^2........(2) ( Rt.Tr. ABJ)

So x = sqrt2.c = y ............(3) (Rt. Isosceles Tr. LKF)

Let < FAJ = p & < FJA = q

Now tan(<LFE) = tan(45-p) = (1 - tan.p)/(1 + tan.p) = (1 - b/(a+b))/(1 + b/(a+b))
So tan(<LFE) = a/(a+2b) = tan.q which yields < LFE = q .....(4)

From (3) & (4), LF and BJ are both equal and parallel
Hence BLFJ is a parallelogram

Sumith Peiris
Moratuwa
Sri Lanka

3. Pure Geometry Solution

Let FAJ = p as before. So < LAB = p
Tr.s LAB & FAD are congruent SAS

Hence < ADF = 135 = < ABL, so L,B,D are collinear
So LB // FJ and also LB = FD = FJ

Hence BLFJ is a parallelogram

Sumith Peiris
Moratuwa
Sri Lanka

4. My solution, if found interesting, can be found at https://stanfulger.blogspot.com/2021/06/problem-1493-gogeometry-my-solution.html

5. BJ^2=b^2+(b+2c)^2=2b^2+4bc+4c^2 while
LF^2=2a^2=2[b+c)^2+c^2]=2b^2+4bc+4c^2
Follows BJ^2=LF^2, BJ=LF
and LB=FD=FJ as required.

6. Can we expect further problems soon, or has this coloumn ended?

1. 7. En la R rotación de centro A y ángulo de 90 grados en sentido antihorario
R(D)=B y R(F)=L.
Lo anterior implica que BL=DF.
8. 