Monday, May 19, 2008

Elearn Geometry Problem 49

See complete Problem 49
Angles, Triangle. Level: High School, SAT Prep, College geometry

1. Let alpha =a. From Tr. ABC, we've Ang EAF = 90 - 3a. EF=AEsin(90-3a)=AEcos(3a). But AB/AE=cos(a). Thus EF=ABcos(3a)cos(a). Hence, (AB+EF)/2 =(ABcos(3a)/cos(a)+AB)/2 =(AB/2)(cos(a)+cos(3a)/cos(a)) =(AB/2)(2cos(a)cos(2a))/cos(a)) = ABcos(2a) or (AB+EF)/2 =ABcos(2a)-----(1). Now from Tr. ABD BD/AB =sin(90-2a) =cos(2a) or BD=ABcos(2a)----(2)
(1) & (2) give us, BD=(AB+EF)/2
Ajit: ajitathle@gmail.com

2. It's possible to solve this problem without using trigonometry.
The key is auxiliary construction:

Geometry problem solving is one of the most challenging skills for students to learn. When a problem requires auxiliary construction, the difficulty of the problem increases drastically, perhaps because deciding which construction to make is an ill-structured problem. By “construction,” we mean adding geometric figures (points, lines, planes) to a problem figure that wasn’t mentioned as "given."

3. AB=AE cos alpha and EF=AE cos 3alpha.Then add the two to get the desired result.

4. Let’a write a for alpha. Take G symmetric of A over the angle C’s bisector, and GH perpendicular to AC.
We have ang(BAD) = 90º-2a, and ang(GAC) = 90º-a, so ang(BAG) = a. That means that AEG is isosceles, with GB = BE.
Thus B is middle point of GE, then GH - BD = BD - EF. But GH = AB by symmetry,
so AB – BD = BD – EF and BD = (AB + EF)/2.

Please let me know what is the language you write in and the corresponding nationality.
(Just curiosity. I'm brazilian and speek portugese).
Thanks.

6. Let AJ be perpendicular to CB, J on BC so that < JAB = 2@. Let AH be the bisector of < JAB, H on BC

So < JAH = < HAB = < BAE = @. Now drop a perpendicular from H to AC, HG with G on AC

Tr. s AHG and ABE are congruent ASA since Tr.s ABE and ABH are congruent ASA

So AB = HG and the result follows since HB = BE

Sumith Peiris
Moratuwa
Sri Lanka

7. http://s11.postimg.org/6wmej6q37/pro_49.png
Draw BG and GH as per sketch
Observe that ∠ (GAB)= alpha and ∠ (GAE)= 2.alpha
Triangle GAE similar to tri. GAC … ( case AA)
So both tri. GAE and GAC are isosceles
In isosceles triangle AGC , heights AB and GH are congruent
In trapezoid GHFE we have BD=1/2(GH+EF)= ½(AB+EF)

8. Reflect AE along AB to AK, triangle ABE = ABK;
area AEC = area ABC- area AEK; area AEC =1/2*EF*AC;
area AKC =area ABC + area ACK =area ABC + area ABE; area AKC=1/2*AB*KC=1/2*AB*AC;
area AEC+ area AKC = 2 area ABC; area ABC=1/2*BD*AC;
so BD=1/2(AB+EF);

9. With the great help of my friend Greg, we have found an original solution.
See original solution on this video

ang(BAC) = 90 - 2α => ang(BAD) = 90 - 2α and ang(ABD) = 2α
Let FJ parallel to BC, J on BD
BJFE is a parallelogram, BJ = EF and ang(DFJ) = 2α
ABE right in B and AEF right in F => A, B, E and F are concyclic, the center O of the circle being the middle of AE
Let BD intersect C in K
ABO isosceles in O => ang(ABO) = α and ang(ABK) = 2α : BO is the angle bisector of ang(ABK) and AB = BK
ang(ABK) = ang(AFK) = ang(DFK) = ang(DFJ) = 2α so on BD J and K are mirror images of one another about AC
So DJ = DK, 2BD = BJ + JD + BK – DK and since BJ = EF and BK = AB then 2BD = EF + AB => BD = (AB+EF )/2
QED.