## Saturday, July 12, 2014

### Geometry Problem 1028: Right Triangle, Double Angle, Midpoint, Congruence

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1028.

1. Solution:
1)Connect B and E Notice that Triangle ABE is isosceles.
2)Let AE = AB = a, let DE = BE = EC = b(right triangle).
3)AEB ~ DBE (They both isosceles and have the same angles). From here we get that AE/BE = BE/DB =
a/b = b/DB --> DB = b^2/a.
4)Draw EN_|_BC and EM _|_AC (notice that EN = DB/2 = b^2/2a).
5)AEM ~ CEN (Same angles) --> AE/EM = EC/ EN = a / EM = b/(b^2/2a).
6)From the proportion in step 5 we get that EM = b/2, in triangle ECM angle ECM = 30 = 90 - 4x.
Solution : The measure of x is 15.

2. Trigonometry solution:
Connect BE and let F is the midpoint of BE
BDE and BAE are isosceles and similar
Let BC= 1we have EC=EB= 1/(2.cos(x)) and BH= 1/(4.cos(x))
And AB= BF/sin(x)= 1/(2.sin(2x))
AC= AB/cos(3x)=BC/sin(3x) => tan(3x)=2. sin(2x)
This equation has 2 solutions: x= 0 and x=15

3. Prob 1027
We build AH perpendicular BC and DF perpendicular AB
CH=HD=1/2DC=1/2ED
DF=DH=1/2ED
α =10
X=90-4α
X=50

Prob 1028

BE=ED=EC EM=EN EN =1/2 EC
Triangle ECN < ECN =30 ECN = 90-4X -> X =15

Prob 1029
We note <BDF = 90
BD perpendicular AC

Erina New Jersey

4. construim AN_|_BE si EM _|_AC, N pe BE,M pe BC,avemAEB ~ DEB(EM=NE=EB=BE/2=CE/2=>ECA=30, 90-4X =30=> X =15

5. Trigonometric Solution 999 and 1028 ( Because they have the same picture and same solution

1. EF perpendicular AB ---> BC=2EF

2. Triangle AEF EF= ABsin2X
3 Triangle ABC BC= ABtg3X

Tg3X=2sin2X From geometric conditions 0 <3X < 90 ---> 0 <X<30

Equation has this solution X=15

Erina New Jersey

6. Hi Antonio

We have 2 other geometric solutions for problems 1028 .
If you want them I can send to you.

Have a great night .

Erina

1. Hi Erina,

7. Let AE = AB = q, BD = r and BE = DE = CE = p

BE is tangential to ADE so p^2 = rq ....(1)

From similar Tr.s EF altitude of Tr. AEF = qr/2p = p/2 from (1)

So < ECA = 30 and hence x = 15

Sumith Peiris
Moratuwa
Sri Lanka