Kristen Robinson

EMAT 4680

3 October 2000

Homework 13 Number 11

In number 11, the exploration was to construct any acute triangle ABC and its circumcircle. Next, construct the three altitudes of ABC, ha, hb, and hc. Third, extend each altitude to its intersection with the circumcircle, forming Ha, Hb, and Hc. The final step was to find (Ha/ha) + (Hb/hb) + (Hc/hc) and prove the result.

It was found that the above sum has a constant value of four. To prove this sum remains constant at this value, we must first analyze other ratios occuring in the exploration.

{[(Ha/ha) + (Ha/(Ha-ha))] + [(Hb/hb) + (Hb/(Hb-hb))] + [(Hc/hc) + (Hc/(Hc-hc))]} -

{[(Ha/(Ha-ha)) - (Ha/ha)] + [(Hb/(Hb-hb)) - (Hb/hb)] + [(Hc/(Hc-hc)) - (Hc/hc)]} = 8.

To simplify:

(Ha/ha) + (Ha/(Ha-ha)) = [Ha(Ha-ha) + Haha]/[ha(Ha-ha)] = Ha^2/(Haha-ha^2),

(Ha/(Ha-ha)) - (Ha/ha) = [Haha - Ha(Ha-ha)]/[ha(Ha-ha)] = (2Haha - Ha^2)/(Haha-ha^2).

So (1) [(Ha/ha) + (Ha/(Ha-ha))] - [(Ha/(Ha-ha)) - (Ha/ha)]

= (Ha^2 - 2Haha + Ha^2)/(Haha-ha^2)

= (2Ha^2 - 2Haha)/(Haha-ha^2) = (2Ha(Ha-ha))/(ha(Ha-ha)) = 2Ha/ha.

Therefore, (2) [(Hb/hb) + (Hb/(Hb-hb))] - [(Hb/(Hb-hb)) - (Hb/hb)] = 2Hb/hb and

(3) [(Hc/hc) + (Hc/(Hc-hc))] - [(Hc/(Hc-hc)) - (Hc/hc)] = 2Hc/hc by the same methods.

Thus, (1) + (2) + (3) = 2Ha/ha + 2Hb/hb + 2Hc/hc = 8.

2(Ha/ha + Hb/hb + Hc/hc) = 8.

Ha/ha + Hb/hb + Hc/hc = 4. This proves the observation. The following figure deomnstrates this scenario.