Slow solution

p94.lisp

@@ -0,0 +1,41 @@
+;; Approach: put one point of the triangle at (0, 0), another at (a,
+;; 0), and the last point at (x, y). Solve for x and then y, and the
+;; area is: (b/4 * sqrt(4a^2 - b^2)). As an optimization, note that
+;; (4a^2 - b^2) must be a perfect square to ensure a rational
+;; solution.
+;;
+;; Note that SQRT is not accurate enough for this
+;; problem. Fortunately, there is ISQRT (which is *not* the same as
+;; (FLOOR (VALUES (SQRT N))) on SBCL, despite what CLHS says).
+;;
+;; Also note that this solution is very slow (takes a minute or so to
+;; compute the answer).
+
+(defun perimeter (a b)
+  "Returns the perimeter of a triangle with two sides of length A and one of length B"
+  (+ a a b))
+
+(defun perfect-square-root (n)
+  "Returns the integer square of of N if N is a perfect square, otherwise returns NIL (note that this function can correctly handle very large integers)"
+  (loop for m upfrom (isqrt n)
+	for square = (square m)
+	until (> square n)
+	do (when (= square n)
+	     (return m))))
+ 
+(defun integer-area-p (a b)
+  "Determines if triangle with two sides of length A and one of length B has an integer area"
+  (let* ((c-squared (- (* 4 a a) (* b b)))
+	 (c (perfect-square-root c-squared)))
+    (when c
+      (integerp (* b 1/4 c)))))
+
+(defun almost-equilateral-triangles (&optional (limit 1000000000))
+  (loop for offset in '(-1 1)
+	sum (loop for a upfrom 2
+		  for b = (+ a offset)
+		  for perimeter = (perimeter a b)
+		  while (<= perimeter limit)
+		  sum (if (integer-area-p a b)
+			  perimeter
+			  0))))