## Wednesday, July 18, 2012

### Problem 786: Right Triangle, Altitude, Hypotenuse, Cevian, Perpendicular, Ratio, Metric relations

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

Click the figure below to see the complete problem 786.

1. http://img7.imageshack.us/img7/479/problem786.png

Draw lines per sketch.
Let AC=3, AH=x => BH=6/5
We have BH^2=AH.HC …( relation in right triangle)
Or 36/25=x(3-x)
We get 2 solutions : x=3/5 or x=12/5
For x=3/5 , HD=2/5 and HE=7/5
So tan(beta)=1/3 and tan(alpha)=7/6
DF/BF=tan(alpha-beta)= 3/5
Due to symmetric , we will get the same result for x=12/5

2. Let AB=a, BC=b, AC=c, BH=h.

Since ab=ch, h/c=2/5,
ab:c^2 = ab:(a^2+b^2) = 2:5

Solving it, we have
a:b = 1:2 or a:b = 2:1

WLOG, let a:b = 1:2.

Let vecBC = 2i, vecBA=j.

Using vector, we have
vecBD=2/3 (i+j)
vecBE=1/3 (4i+j)

Therefore, angle DBC = 45 degrees
also, tan(angle EBC)=1/4

Hence,
DF/BF
= tan(angle DBE)
= tan(45 - angle EBC)
= [1 - 1/4] / [1 + 1/4]
= 3/5

3. let AC=30, then BH=12
AB=6sqrt(5), BC=12sqrt(5), AH=6,HD=4, BD=4sqrt(10), BE=6sqrt(10)
the area of triangle BDE is 60
then DF=2sqrt(10)
DF/BF=sqrt(3)/3

1. To Anonymous (Problem 786) your solution is not correct.

4. Let AH = p, BD = d, BE = e and BF = h and BF = g

AH. HC = BH^2 so

p(b-p) = 4b^2/25

Solving the quadratic p = b/5 or 4b/5

Take p = b/5; c^2 = b^2/5 and a^2 = 4b^2/5

Hence a/c = 2 and so BD bisects the right angle at B

HD = b/3 -b/5 = 2b/15 so d^2 = 8b^2/45

Similarly from right Tr. BHE e^2 = 17b^2/45

From the calculations using pure geometry alone become a bit unwieldy

We can use the fact that eh = 2b^2/15 by noting that the Tr.s ABD and EBD have equal areas

Also since BFDH is cyclic h/(2b/5) = (b/3)/e = (e-g)/ (2b/3)

But the easiest way is to use Trigonometry as Jacob has done above so that h/g = tan DBF = (1-1/3)/

5. Correction tanDBF = (1-1/4)/(1+1/4)= 3/5