Find the smallest obtuse triangle such that its sides, the altitude to the obtuse angle, and the median to the largest side are all integers.
In case it arises, "smallest" refers to the area of the triangle.
Computer trial and error leads to a triangle with sides 25, 39 and 56, with a median of 17 and altitude of 15.
The display also shows the angle in degrees on the right side of the first line:
25 39 56 24.22775 9.10460 146.66765
17 15
There exist larger such triangles, but the following display, with two rows per triangle, shows also some spurious results, that have almost, but not quite, integral altitudes:
sides angles (degrees)
median altitude
50 78 112 24.22775 9.10460 146.66765
33.99999999999999 30
75 117 168 24.22775 9.10460 146.66765
50.99999999999999 44.99999999999999
100 156 224 24.22775 9.10460 146.66765
67.99999999999999 59.99999999999999
125 195 280 24.22775 9.10460 146.66765
84.99999999999999 74.99999999999999
150 234 336 24.22775 9.10460 146.66765
102 89.99999999999999
139 327 370 58.11546 8.53232 113.35222
170 121.9999995210097
175 273 392 24.22775 9.10460 146.66765
119 105
212 238 390 18.19679 14.08910 147.71411
113 112
200 312 448 24.22775 9.10460 146.66765
136 120
105 445 500 46.84761 2.07802 131.07437
205 84
225 351 504 24.22775 9.10460 146.66765
153 135
250 390 560 24.22775 9.10460 146.66765
170 150
275 429 616 24.22775 9.10460 146.66765
187 165
300 468 672 24.22775 9.10460 146.66765
204 180
325 507 728 24.22775 9.10460 146.66765
221 195
278 654 740 58.11546 8.53232 113.35222
340 243.9999990420194
232 650 798 33.29749 3.57721 143.12531
281 160
350 546 784 24.22775 9.10460 146.66765
238 210
424 476 780 18.19679 14.08910 147.71411
226 224
375 585 840 24.22775 9.10460 146.66765
255 225
400 624 896 24.22775 9.10460 146.66765
271.9999999999999 240
241 777 952 25.93495 2.14578 151.91927
323 148.999999763032
Many of the above are integral multiples of the smallest such triangle. Those altitudes where the fractional part consists of all 9's are really integers and the amount off is due to rounding errors. But the ones where the last several places after the decimal are not 9 are really only close to integral in value. I can theorize that since we are varying a total of 9 digits in the three side lengths, that enough triangles exist that probabilistically speaking, some would have altitudes close to, but not, integers.
The smallest triangles of a given shape (defining a set of similar triangles) are:
25 39 56 24.22775 9.10460 146.66765
17 15
212 238 390 18.19679 14.08910 147.71411
113 112
105 445 500 46.84761 2.07802 131.07437
205 84
232 650 798 33.29749 3.57721 143.12531
281 160
... in ascending order of perimeter. By area, the middle two would be switched.
The list probably continues, but the trial and error takes more time as the sizes get larger.
DEFDBL A-Z
pi = ATN(1) * 4
CLS
FOR t = 4 TO 1000000
FOR s1 = 1 TO INT(t / 3)
FOR s2 = s1 + 1 TO INT((t - s1) / 2)
s3 = t - s1 - s2
IF s3 * s3 > s1 * s1 + s2 * s2 AND s3 < s1 + s2 THEN
cosA = (s3 * s3 + s1 * s1 - s2 * s2) / (2 * s3 * s1)
b = s3 / 2
med = SQR(s1 * s1 + b * b - 2 * b * s1 * cosA)
medi = INT(med + .5)
IF ABS(med - medi) / med < .00000001# THEN
alt = s1 * SQR(1 - cosA * cosA)
alti = INT(alt + .5)
IF ABS(alt - alti) / alt < .00000001# THEN
cosC = (s3 * s3 + s2 * s2 - s1 * s1) / (2 * s3 * s2)
A = ATN((1 - cosA * cosA) / cosA) * 180 / pi
C = ATN((1 - cosC * cosC) / cosC) * 180 / pi
PRINT s1; s2; s3; TAB(45);
PRINT USING " ###.#####"; A; C; 180 - A - C
PRINT med; TAB(40); alt
ct = ct + 1
IF ct > 22 THEN END
END IF
END IF
END IF
NEXT
NEXT
NEXT
so I did not have enough zeros in .00000001# to weed out the spurious values shown.
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Posted by Charlie
on 2006-05-05 15:32:40 |