<p> Triangle in a Triangle</p><p>Part 1</p><p>Y β</p><p>Δ 2 c a</p><p>Δ 4 Δ 1 Δ 3 Z α γ X b</p><p>Let the area of triangle XYZ be Δ, and the smaller triangles be Δ1 through Δ4, as above.</p><p>1 The area of the large triangle can be found as bc sin   . 2</p><p>The proportions into which each side of the large triangle has been split means that, the base 1 2 of Δ is the length of the base of the large triangle, and the sloping side that of the 1 3 3 1 b 2c 2  1  2 large triangle. So the area of Δ1 is sin   bc sin    . The same holds for 2 3 3 9  2  9 2 triangles 2 and 3. So        1 2 3 9</p><p>2 1 The area of the inner triangle, triangle 4, is therefore   3     . 9 3</p><p>Page 1 of 4 Part 2</p><p>X</p><p>U</p><p> a B</p><p> x Δ Z 2 α C A y O Y W b</p><p>Ratio of Interior Lines</p><p>Essentially from the hints section:</p><p>Let the vectors OX  a and OY  b . Therefore YX  a  b .</p><p>1 1 This means that OZ  OY YZ  b  a  b   a  2b , and that 3 3 b 1 WX WO  OX    a  3a  b . 3 3</p><p> The displacement to any point along OZ can be written as a  2b , and any point along 3 b  1 WX can be written as OW  WX   3a  b   3a  b 1  . The point of 3 3 3 intersection, A, will be when both displacements along OZ and WX are equal.</p><p>a 2b b1   Equating the displacements gives the equations  a and  , or written 3 3 3 more simply:   3 and 2  1   . Substituting from the first into the second gives: 1 3 6  1   , or that   , and so   . 7 7</p><p>This means that the point of intersection, A, is 1/7 the way along WX from W, and 3/7 the way along OZ from O. The same holds for B, and C, and so each of the lines is split into the ratio 3:3:1 from the vertex, by the other lines.</p><p>Area of the Interior Triangle</p><p>Let the area of triangle OXY be Δ. Let the area of the inner triangle be Δ2 as in the diagram. Using the cross, or vector, product, the area of triangle OXY can be found as 1 1 OX OY  a  b   . 2 2</p><p>Page 2 of 4 3 3 From the previous part AB  WX , and AC  OZ . And so 7 7 3 1 1 3 1 1 AB  3a  b   3a  b , and AC  a  2b   a  2b  . 7 3 7 7 3 7</p><p>1 Using the vector product again, the area of the inner triangle, Δ2, is equal to AB  AC , this 2 then can be found as 1 1 1 1 1 1 3a  a  3a  2b  b  a  2b  b 3a  b  a  2b   3a  b  a  2b   . A 2 7 7 2 49 2 49 vector crossed with itself is 0, and remembering that u v  v u , this can be simplified as</p><p>1 6a  b  a  b 1 7a  b 1 a  b 1  1  1     a  b    , and so the area of the interior 2 49 2 49 2 7 7  2  7 triangle, ABC, is 1/7 the area of the original triangle.</p><p>Area of the Small Triangles at the Edges</p><p>Concentrating on triangle ABC, and triangle OAW for the moment.</p><p>3 3 1 From the first part OA  AC  OZ , AB  WX , and WA  WX , which means that 7 7 7 1 WA  AB . Let the length of AB be x, and the length of AC be y, as in the diagram. 3</p><p>OAW  BAC because the two intersecting lines are straight. Let the BAC   , as in the diagram.</p><p>1 The area of triangle ABC could be found as xy sin . Then the area of triangle OAW is 2 1  x  1  1   y sin   xy sin  , so the area of triangle OAW is 1/3 the area of the interior 2  3  3  2  triangle. Remembering that the interior triangle is 1/7 the area of the original triangle, the 1 1 1 area of OAW is then    , or 1/21 the area of the original triangle. 3 7 21</p><p>The same argument applies to each of the other small edge triangles, XBU, and YZC. Each edge triangle is 1/21 the area of the original triangle, and so each edge triangle has the same area the others.</p><p>Area of the Quadrilaterals</p><p>Concentrating on triangle OWX for the moment.</p><p>Triangle OWX has the same height as triangle OXY, but the base of OWX is 1/3 that of OXY. This means that the area of OWX is equal to 1/3 the area of triangle OXY.</p><p>Page 3 of 4 The two small edge triangles each have area 1/21 that of OXY. The difference between this and the area of OWX gives the area of quadrilateral OUBA. The area is found to be   7  2 5  2      . The same argument applies to each of the other quadrilaterals, 3 21 21 21 and so the area of each quadrilateral is 5/21 the area of the original triangle, and so each quadrilateral has the same area the others.</p><p>Original Area</p><p>As a quick check, adding the areas of the quadrilaterals, interior, and edge triangles produces 5 1 1 15  3  3  3    3          . The areas of the triangles and quadrilaterals 21 21 7  21  do indeed add up to the area of the original triangle.</p><p>Page 4 of 4</p>