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diff --git a/challenge-187/jo-37/perl/ch-2.pl b/challenge-187/jo-37/perl/ch-2.pl
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+#!/usr/bin/perl -s
+
+use v5.16;
+use Test2::V0;
+use experimental 'signatures';
+
+our ($tests, $examples);
+
+run_tests() if $tests || $examples;	# does not return
+
+die <<EOS unless @ARGV;
+usage: $0 [-examples] [-tests] [N...]
+
+-examples
+    run the examples from the challenge
+ 
+-tests
+    run some tests
+
+N...
+    List of numbers.
+
+EOS
+
+
+### Input and Output
+
+main: {
+    local $" = ', ';
+    say "(@{[mp_triangle(@ARGV)]})";
+}
+
+
+### Implementation
+
+# Demystifying the task: There is no magic in triples satisfying the
+# given inequalities.  These are just edges that form a non-degenerated
+# triangle.  The task may be formulated as finding the triangle having a
+# maximum perimeter from a given set of edges.
+# Note: There is no triangle in any sequence growing at least like a
+# Fibonacci sequence.
+sub mp_triangle {
+    # Sort descending.
+	my @e = sort {$b <=> $a} @_;
+    while () {
+        # There is no triangle, if there is no smallest edge having
+        # a positive length.
+        return () unless ($e[2] // 0) > 0;
+        # If the largest three numbers form a triangle, it has a maximum
+        # perimeter.
+        return @e[0, 1, 2] if $e[0] < $e[1] + $e[2];
+        # If the largest three edges do not form a triangle then there
+        # is no triangle including the largest edge at all.  Drop it.
+        shift @e;
+    }
+}
+
+
+### Examples and tests
+
+sub run_tests {
+    SKIP: {
+        skip "examples" unless $examples;
+
+        is [mp_triangle(1, 2, 3, 2)], [3, 2, 2], 'example 1';
+        is [mp_triangle(1, 3, 2)], [], 'example 2';
+        is [mp_triangle(1, 1, 2, 3)], [], 'example 3';
+        is [mp_triangle(2, 4, 3)], [4, 3, 2], 'example 4';
+    }
+
+    SKIP: {
+        skip "tests" unless $tests;
+
+        is [mp_triangle(1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144)], [], 
+            'no triangle from Fibonacci numbers';
+
+        is [mp_triangle(1, 2, 2, 3, 4, 6, 9, 14, 22, 35)],
+            [35, 22, 14], 'F(N) = F(N-2) + F(N-1) - 1';
+	}
+
+    done_testing;
+    exit;
+}