All About Ratios

5511 /www/nrich/html/content/id/5511/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Can you work out the distances? <mdo:image alt="the problem" height="478" src="AllAboutRatios.GIF" width="611"></mdo:image> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Excellent solutions were sent in by Michael from Wilmslow High School, Cheshire, UK and Andrei from Tudor Vianu National College, Bucharest, Romania. <table border="0"> <tbody> <tr> <td><mdo:image height="150" width="322" src="ratio1.jpg" alt="ratio1"></mdo:image></td> <td>(1) As triangles AOD and AEC are similar $${OA\over AC} = {OD\over EC} = {2\over 3} {\Rightarrow}{OA\over OC} = {2\over 5}.$$ As triangles BOF and BCG are similar $${OB\over BC} = {OF\over GC} = {3\over 1} {\Rightarrow} {OB\over OC} = {3\over 4}.$$ Hence: $${AB\over OC} = {OB - OA\over OC} = {3\over 4} - {2\over 5} = {7\over 20}.$$</td> </tr> </tbody> </table> <table border="0"> <tbody> <tr> <td><mdo:image height="180" width="322" src="ratio2.jpg" alt="ratio2"></mdo:image></td> <td>(2) From triangles OAD and AXE $${OA\over AX} = {OD\over EX} = {2\over 1.5}= {4\over 3}$$ and hence $${OA\over OX} = {4\over 7} \quad {\rm and} \quad {OA\over OC} = {2\over 7}.$$ Again $${OB\over OC} = {3\over 4}.$$ Hence: $${AB\over OC} = {OB - OA\over OC} = {3\over 4} - {2\over 7} = {13\over 28}.$$</td> </tr> </tbody> </table> <a href="" class="control"></a> <a class="control" href=""></a> <table border="0"> <tbody> <tr> <td><mdo:image height="180" width="320" src="ratio3.jpg" alt="ratio3"></mdo:image></td> <td>(3) The reasoning in this part is identical to part (1).</td> </tr> </tbody> </table> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Finding several methods can often help us to a better understanding of the mathematics involved, giving us a richer experience than gained from simply finding the answer. Although these ratios can be found using coordinates and finding intersections of lines and distances, this is not the most efficient method. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">There is a better method using only ratios. Look for similar triangles. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> (1) By similar triangles $${OB\over BC}= {3\over 1}{\rm so} {OB\over OC}={3\over 4}.$$ Again by similar triangles, $${OA\over OC}= {2\over 3} {\rm so} {OA\over AC}={2\over 5}.$$ The ratio $${AB\over OC} = {OB - OA\over OC} = {3\over 4} - {2\over 5} = {7\over 20}.$$ (2) By similar triangles $${OB\over BC}= {3\over 1}{\rm so} {OB\over OC}={3\over 4}.$$ Again by similar triangles, $${OA\over OX}= {4\over 3}$$ where $X$ is the midpoint of $OC$ so $${OA\over OC}={4\over 14}.$$ The ratio $${AB\over OC} = {OB - OA\over OC} = {3\over 4} - {2\over 7} = {13\over 28}.$$ (3) This is exactly the same as (1) By similar triangles $${OB\over BC}= {3\over 1} {\rm so} {OB\over OC}={3\over 4}.$$ Again by similar triangles, $${OA\over OC}= {2\over 3} {\rm so} {OA\over AC}={2\over 5}.$$ The ratio $${AB\over OC} = {OB - OA\over OC} = {3\over 4} - {2\over 5} = {7\over 20}.$$ Alternatively: (1) $AB = {7\over 20}\sqrt{10}$, $OC = \sqrt{10}$. (2) $AB ={13 \sqrt 13\over 28}$, $OC = \sqrt 13$. (3) $AB ={7\sqrt 13\over 20}$, $OC = \sqrt 13$. </mdoxml> 2 3 0 0 0 0 1 All About Ratios A new problem posed by Lyndon Baker who has devised many NRICH problems over the years. 2D Geometry, Shape and Space Similar triangles Coordinates and Coordinate Geometry Dividing line in given ratio Fractions, Decimals, Percentages, Ratio and Proportion Ratio