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TchrWill
Advanced Member

USA
79 Posts

Posted - 03/25/2013 :  10:56:04  Show Profile  Reply with Quote
What is the smallest triangle possible with all sides and altitudes being six different integers?
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someguy
Advanced Member

Canada
143 Posts

Posted - 03/26/2013 :  01:08:36  Show Profile  Reply with Quote
Edge lengths: 15, 20, 25
Perimeter: 60
Area: 150
Not sure what your measure you are using to determine size,
but this is the one with smallest perimeter and area.

If you want the smallest one that is not a right triangle,
then it is the one that follows.
Edge lengths: 35, 75, 100
Perimeter: 210
Area: 1050

My approach was based on the observation that if you look at the altitude for the longest side, it partitions the triangle into two right triangles. It's not too hard at this stage to just search for solutions based on the lengths of this altitude and the lengths of the two other sides (one is longer than the other and both are longer than the altitude). A computer helps here.
Instead, I noted that the two right triangles must both have sides of integer length (the sum of the sides that make up the longest side of the initial triangle is an integer so the lengths of the two sides of the right triangles that create it can't be irrational which also implies that they must be integers since the other side lengths are integers). Anyway, knowing how to generate primitive Pythagorean triples and then scaling pairs to match up a side length allows for calculations and checking that can be done by hand if one desires.
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TchrWill
Advanced Member

USA
79 Posts

Posted - 03/27/2013 :  10:10:15  Show Profile  Reply with Quote
Nice approach someguy. As usual, I meander down the algebraic path. It more often than not, takes more time but eventually I get there.

1--Create triangle ABC, A at the top, B bottom right and C bottom left, angle ACD smaller than angle ABC.
2--Draw altitude AD = ha from A to D on side a.
3--Let CD = a1, DB = a2, AC = b and AB = c.
4--a1 = sqrt(b^2 - ha^2) and a2 = sqrt(c^2 - ha^2)
5--a1 and a2 must be integers.
6--The sides of integral Pythagorean triangles derive from x = a^2 - b^2, y = 2ab, and z = a^2 + b^2 where a and b are two positive integers, a greater than b.
7--The smallest possible integral right triangle derives from a = 2 and b = 1 which p
Whew!roduces the ever familiar 3-4-5 right triangle.
8--Assume that the two right triangles created, ADC and ADB, are from this family, i.e., multiples of the 3-4-5 triangle (which means that the whole triangle is of the same family) to derive the smallest area triangle.
9--Letting m and n be the multipliers of ADB and ADC respectively, ha = 4m = 3n, b = 5n, c = 5m, a1 = 4n, a2 = 3m.
10--m = 3 and n = 4 produce the smallest ha = 12 making a = 25, b = 20, and c = 15.
11--So we have a right triangle of the 3-4-5 family.
12--But, this has only 4 different integers for the sides and altitudes.
13--So the triangle we seek must be obtuse.
14--Redrawing our triangle ABC, A at the top, B bottom right, C bottom right but closer to A.
15--Draw a vertical line from A to D on BC extended to D. AD = ha = the altitude to side BC.
16--Triangle ABC is the one we are seeking the integers for but we will use triangles ABD and ACD to find it.
17--Let AC = b, AB = c, CD = a1 and BC = a2.
18--Since both triangles ACD and ABD must be Pythagorean integer triangles, we will again assume that both are of the 3-4-5 (3-4-5 being the smallest integral Pythagorean triangle) family in order to seek out the minimum area.
19--Letting m and n be the multipliers for ACD and ABC respectively, ha = 4m = 3n, b = 5m, c = 5n, a1 = 3m and (a1 + a2) = 4n, a2 = 4n - 3m.
20--The area of triangle ABC is A = (4n - 3m)4m/2 = 8mn - 6m^2 = hc(5n)/2 = hb(5m)/2..
21--m = 3 and n = 4 to create the smallest ha = 4m = 3n = 12.
22--Then, hb = 2A/5m = (16mn - 12m^2)/5m and hc = (16mn - 12m^2)/5
23--Substituting m = 3 and n = 4, hb = 28/5 and hc = 21/5.
24--Therefore, we have ha = 4m = 3n = 12, a2 = 4n - 3m = 7, b = 5m = 15, c = 5n = 20, hb = 28/5 and hc = 21/5.
25--In order to rid ourselves of the two fractional altitudes, we need to multiply all the sides and altitudes by 5 giving us
a = 35, b = 75, c = 100, ha = 60, hb = 28 and hc = 21 and an area of triangle ABC of 1050.

Whew!

You seem to be the only one interested in problems/puzzles of this kind. I hope others will join in.

Edited by - TchrWill on 03/29/2013 08:31:05
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someguy
Advanced Member

Canada
143 Posts

Posted - 03/28/2013 :  17:06:39  Show Profile  Reply with Quote
Hi TchrWill,

Our approaches are very similar. We both start with right triangles having edge lengths that form sets of primitive Pythagorean triples. We both scale these triangles to get edges of common lengths so we can perform our construction, and then scale the result again to make the altitudes integers. They differ in that I join the two triangles to make a larger one, you remove a smaller right triangle from the larger to create the desired triangle.

My initial attempt was very similar to yours, but since we are scaling our triangles twice, I was concerned about whether or not we could start with larger initial triangles that need to be scaled less resulting in a smaller final triangle. Because of this, I also checked other pairs of starting triangle (each pair can be joined together in 4 ways, but we don't have to check that many pairs since the area of the small one we found was 1050).

Thanks for pointing out that with a right triangle, the 6 integers are not all distinct. My thought when I wrote that was that the sides lengths are all different so the altitudes will all be different. I neglected to consider that the altitudes and sides were the same. Of course, with a right triangle, the legs are also altitudes! (silly me ).

You might want to fix the typo in steps 18 and 19 in your post (change ABC to ABD since the right triangles are ABD and ACD in your construction).


I look forward to your next one if you have more. Given the number of times your posts get viewed, it seems reasonable that others are checking them out. If you do post another, I will hold off from posting my thoughts for a few days. Hopefully that will allow someone else to get involved.
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