Shortcuts

 Sitemap Contact Newsletter Store Books Features Gallery E-cards Games

 ••• Related links Puzzles workshops for schools & museums. Editorial content and syndication puzzles for the media, editors & publishers. Abracada-Brain Teasers and Puzzles! Numbers, just numbers... Have a Math question? Ask Dr. Math! ••• ••• Logic smile! Did you hear the one about the patient who forgot to take his homeopathic medicine? He died of an overdose!       - Unknown ••• SPACE INVADERS Play Space Invaders online!

# Previous Puzzles of the Month + Solutions

July-August 2008, Puzzle nr 118
Back to Puzzle-of-the-Month page | Home

Puzzle # 118 Difficulty level: , general math knowledge.

 Ratio of similar triangles.   If the segment A'B' is tangent to the incircle of triangle ABC, and that segment AB = segment CM; then, what is the ratio of the area of the triangle ABC to the area of the small triangle A'B’C? Hint: angle CÂB is not necessarily a right angle, however triangles ABC and A'B'C are similar! Keywords: inscribed circle, incircle, tangent. Related puzzles: - Soccer ball problem. - Prof. Gibbus' angle. - Triangoli proporzionevoli Se il segmento A'B' è tangente al cerchio inscritto nel triangolo ABC dell'immagine qui sotto, e che AB = CM; quant’è grande, allora, il triangolo ABC rispetto al piccolo triangolo A’B’C in termini di superficie? Piccolo indizio: l'angolo CÂB non è per forza un angolo retto, tuttavia i triangoli ABC e A'B'C sono simili! Parole chiave: cerchio inscritto, tangente. Suggerisci un'altra soluzione Chiudi - Triangles proportionnels... Si le segment A'B' est tangent au cercle inscrit du triangle ABC illustré ci-dessous, et que AB = CM; quel est alors le rapport de grandeur du triangle ABC au petit triangle A’B’C en terme de surface? Une piste: l'angle CÂB n'est pas obligatoirement un angle droit, toutefois les triangles ABC et A'B'C sont sembables! Mots clés: cercle inscrit, tangente. Propose une autre solution Fermer Source of the puzzle: ©G. Sarcone, "Focus Braintrainer Magazine #16", page 61.

 More Math Facts behind the puzzle Let the circle (O, OM) be the incircle of the triangle ABC above. Consider that: AB = c BC = a AC = b AL = AN = x BL = BM = y CM = CN = z Semiperimeter p = 1/2(a + b + c) Then: 2x = (x + y) + (x + z) - (y + z) 2x = AB + AC - BC and x = 1/2(AB + AC - BC) x = 1/2(AB + AC + BC) - BC x = p - a Furthermore: y = p - b z = p - c Formulae Area A of the triangle ABC: A = (px · py · pz) = [p(p - a)(p - b)(p - c)] = r · p A = 1/2(bc · sinα) = 1/2(ac · sinβ) = 1/2(ab · sinγ)

Back to Puzzle-of-the-Month page | Home