perplexus.info :: Geometry : Two Triangles Divided

Two Triangles Divided (Posted on 2007-12-16)

There are two triangles ABC, and A'B'C'. The bases are AB and A'B'.

Triangle ABC has an altitude drawn from C to AB, meeting the base at point P. This altitude divides the triangle into two unequal right triangles.

Triangle A'B'C' also has a point, P' on its base, with a line segment connecting it to vertex C', but chosen so that angle A'P'C' is 60°, with the resulting triangle A'P'C' though, not being equilateral.

All eight line segments are of integer length, and each triangle has a perimeter less than 50. The bases, AB and A'B', are the longest sides in each of the two respective original triangles, and they differ by 1 unit in length.

What are the dimensions of the triangles ABC and A'B'C'?
What are the lengths of CP and C'P'?

Submitted by Charlie

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Solution: (Hide)

Only one set of dimensions works for triangle ABC: base AC BC CP AP perimeter 21 10 17 8 6 48 So the dimensions of ABC are: base 21, sides 10 and 17. Several work for A'B'C': 11 7 7 5 8 25 13 7 7 3 8 27 22 13 13 8 15 48 23 13 13 7 15 49

but in only one instance does the base differ from 21 by 1 unit, in bold above. So the dimensions of A'B'C' are: base 22, sides both 13.
CP and C'P' are each 8 units long.

DEFDBL A-Z CLS FOR peri = 4 TO 49 FOR bse = 1 TO -INT(-peri / 2) - 1 remain = peri - bse low = -INT(-(remain - bse) / 2) FOR s1 = low TO remain / 2 s2 = remain - s1 cA = (s1 * s1 + bse * bse - s2 * s2) / (2 * s1 * bse) sA = SQR(1 - cA * cA) bdiv = cA * s1 ibdiv = INT(bdiv + .5) IF ABS(bdiv - ibdiv) < 1E-08 AND s1 <> s2 THEN alt = sA * s1 ialt = INT(alt + .5) IF ABS(alt - ialt) < 1E-08 AND ialt > 0 AND s1 <= bse AND s2 <= bse THEN PRINT USING \\"########\\"; bse; s1; s2; ialt; ibdiv; peri END IF END IF NEXT NEXT NEXT PRINT FOR peri = 4 TO 49 FOR bse = 1 TO -INT(-peri / 2) - 1 remain = peri - bse low = -INT(-(remain - bse) / 2) FOR s1 = low TO remain / 2 s2 = remain - s1 cA = (s1 * s1 + bse * bse - s2 * s2) / (2 * s1 * bse) sA = SQR(1 - cA * cA) FOR bdiv = 1 TO bse - 1 sidea = SQR(s1 * s1 + bdiv * bdiv - 2 * s1 * bdiv * cA) isidea = INT(sidea + .5) IF ABS(sidea - isidea) < 1E-08 AND s1 <= bse AND s2 <= bse THEN sP = sA * s1 / sidea IF ABS(sP - SQR(3) / 2) < 1E-08 AND bdiv > cA * s1 THEN IF ABS(sA - SQR(3) / 2) > 1E-08 THEN PRINT USING "########"; bse; s1; s2; isidea; bdiv; peri END IF END IF END IF NEXT NEXT NEXT NEXT

From Enigma No. 1467 by Richard England, New Scientist, 3 November 2007

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: Answer (Solution)	DJ	2007-12-17 21:45:11
Solution	Dej Mar	2007-12-17 21:03:37
Answer	DJ	2007-12-17 17:30:26
Partial Solution	Praneeth	2007-12-17 01:43:57

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