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 Tidy Triangle (Posted on 2006-05-05)
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.

 See The Solution Submitted by Jer No Rating

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 computer trial-and-error solution (spoiler) | Comment 1 of 7

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  146.6676517                                     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  146.6676533.99999999999999                      30`
`75  117  168                                  24.22775    9.10460  146.6676550.99999999999999                      44.99999999999999`
`100  156  224                                 24.22775    9.10460  146.6676567.99999999999999                      59.99999999999999`
`125  195  280                                 24.22775    9.10460  146.6676584.99999999999999                      74.99999999999999`
`150  234  336                                 24.22775    9.10460  146.66765102                                    89.99999999999999`
`139  327  370                                 58.11546    8.53232  113.35222170                                    121.9999995210097`
`175  273  392                                 24.22775    9.10460  146.66765119                                    105`
`212  238  390                                 18.19679   14.08910  147.71411113                                    112`
`200  312  448                                 24.22775    9.10460  146.66765136                                    120`
`105  445  500                                 46.84761    2.07802  131.07437205                                    84`
`225  351  504                                 24.22775    9.10460  146.66765153                                    135`
`250  390  560                                 24.22775    9.10460  146.66765170                                    150`
`275  429  616                                 24.22775    9.10460  146.66765187                                    165`
`300  468  672                                 24.22775    9.10460  146.66765204                                    180`
`325  507  728                                 24.22775    9.10460  146.66765221                                    195`
`278  654  740                                 58.11546    8.53232  113.35222340                                    243.9999990420194`
`232  650  798                                 33.29749    3.57721  143.12531281                                    160`
`350  546  784                                 24.22775    9.10460  146.66765238                                    210`
`424  476  780                                 18.19679   14.08910  147.71411226                                    224`
`375  585  840                                 24.22775    9.10460  146.66765255                                    225`
`400  624  896                                 24.22775    9.10460  146.66765271.9999999999999                      240`
`241  777  952                                 25.93495    2.14578  151.91927323                                    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  146.6676517                                     15`
`212  238  390                                 18.19679   14.08910  147.71411113                                    112`
`105  445  500                                 46.84761    2.07802  131.07437205                                    84`
`232  650  798                                 33.29749    3.57721  143.12531281                                    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.

 Posted by Charlie on 2006-05-05 15:32:40
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