## Monday, March 10, 2008

### A 5-7-8 Triangle for Starters -- An In-Depth Exploration in Algebra, Geometry and Trigonometry

As promised, here is an investigation/activity/challenge/.../ (Anyone remember a certain star college and pro football player, with initials K.S., who was nicknamed 'Slash' because he could play several positions?).

The following extensive activity is designed as a long-term assignment, however, modify it as you wish. Don't forget to forgive proper attribution as indicated in the sidebar.

A Possible Instructional Scenario...
"Girls and boys, what do you think of when you're told that a triangle has a 60° angle? What if you're given that all the sides have integer lengths? Why can't it be 30-60-90 in that case? Of course, it could still be equilateral, but for this exploration, we are looking for scalene triangles with integer sides and a 60° angle. By the way, if it's not equilateral, how do we know it must be scalene? Why couldn't it be isosceles?"
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The Exploration

Let's agree on the following labeling for our triangle:

ΔABC with ∠C = 60° with opposite side c. Other angles are labeled A and B with opposite sides a and b, respectively. For this investigation, b > a.

(a) Show that the angle opposite the '7' side in the 5-7-8 triangle in the title of this post is 60°. Using the labeling above: If a=5, b=8, c=7, show that ∠C = 60°.

Note: This can be done with or without the Law of Cosines. As Joshua pointed out in a previous post, the student can work with the altitude on the '5' side and use the Pythagorean Theorem and algebra to show that a 30-60-90 triangle is formed. This method is instructive (and constructive too!).

For parts (b) - (e), we will no longer be focusing only on the 5-7-8 triangle. Consider any integer-sided ΔABC with ∠C = 60° and with opposite side of length 7.
(b) Show that a2 + b2 - ab = 49. [*]
See note after part (a) for two possible methods.

(c) We are looking for integer solutions to [*]. There are several methods including a Pell equation approach (we will not go in that direction). Students can certainly try a guess-test strategy, however we can initially restrict the possible values of b, can't we?
As an initial boundary, show that 13 >b ≥ 8 using basic geometry.

(d) Show that a better restriction for b is 14/√3 b ≥ 8 by:

(i) [Trig] Using the Law of Sines
(ii) [Advanced Algebra] Use the quadratic formula in [*] to solve for a in terms of b. Using the discriminant, show that b ≤ 14/√3.

From this result, explain why it follows that b must equal 8.

(e) Solve for a.
Suggestions: From [*] OR use your result from the quadratic formula in (d)(ii) to show that there are exactly 2 scalene triangles satisfying the above conditions.
a=3, b=8, c=7;
a=5, b=8, c=7
BTW, is it a coincidence that the two values of a happen to add up to b?

(f) REPEAT PARTS (b) - (e) for an integer-sided scalene triangle with a 60° angle and an opposite side of length 13. There will be some slight modifications needed such as formula [*]. State and use a similar inequality from (d) to show that there are two possible values for b, namely 14 and 15. Go further and show b=14 is not possible (e.g., using the discriminant).
Then show that b = 15 leads to two solutions (triangles) - sorry, I'm not giving these away yet!

(g) REPEAT PARTS (b) - (e) for an integer-sided scalene triangle with a 60° angle and an opposite side of length 19. Again, you should find that the inequality from (d) leads to two possible values for b, only one of which works. Give the two solutions (triangles).

(h) Some of you will no doubt wonder why we did 3 separate analyses, when a slightly more general approach could have been used, specifically a more general inequality in part (d). Ok, so do that (keep all conditions, except side 'c' will now be a parameter).
Show that, in general, 2c/√3 b ≥ c.

(j) Surely, we can't go further other than searching for a general solution for the 60° problem. Of course, we can: Come up with similar questions and solutions to all parts above if ∠C = 120°. I'll start you off: a=3, b=5, c=7. Show that ∠C = 120°, etc...

I'm sure our astute and talented readers will pick up on my usual errors or omissions or make suggestions to improve the flow of the activity. I'm counting on you! Enjoy...