Summer students' sandbox

8461 /www/nrich/html/content/id/8461/ 3 0000-00-00T00:00:00 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This is the solution by Tim from Gravesend Grammar School. Assume that two of the internal angle bisectors, $AM$ and $BN$, are perpendicular to each other, meeting at $X$, ie $\angle AXB = 90^\circ$. In triangle $\Delta AXB$, $\angle AXB + \angle BAX + \angle XBA = 180^\circ$ so $\angle BAX + \angle XBA = 90^\circ$. But $\angle XAC = \angle XAB$ and $\angle ABX = \angle XBC$ so the sum of the angles in the triangle is $2 ( \angle BAX + \angle XBA ) + \angle BCA = 180^\circ + \angle BCA$ so $\angle BCA = 0$, so $ABC$ is not a triangle as it only has two angles, hence $AM$ and $BN$ are not perpendicular. Note: What is happening here is that $BC$ is parallel to $AC$. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><table style="" border=""> <tbody> <tr> <td valign="top" width="300" style=""><mdo:image align="bottom" alt="*" border="0" src="triangle.gif"></mdo:image> A'B'C' is an equilateral triangle. For any point P inside it, the area of <mdo:image align="bottom" alt="*" border="0" src="triangle.gif"></mdo:image> A'B'C' is equal to the sum of the areas of the triangles PB'C' , PC'A' and PA'B' PA, PB and PC are the heights of triangles PB'C', PC'A' and PA'B'. If the length of the side of the equilateral triangle is L units, then: Area <mdo:image align="bottom" alt="*" border="0" src="triangle.gif"></mdo:image> A'B'C' = ½ L x (PA + PB + PC) PA + PB + PC = (2 x Area <mdo:image align="bottom" alt="*" border="0" src="triangle.gif"></mdo:image> A'B'C')/L = ½L <mdo:image align="bottom" alt="*" border="0" src="root.gif"></mdo:image> 3 = constant. </td> <td valign="top" width="290" style=""><mdo:image alt="Figure 1" src="sept_fig1.gif"></mdo:image></td> </tr> <tr> <td colspan="2" width="590" style="">For each position of the point P inside the equilateral triangle we can take PA, PB and PC to represent the percentages of something which is split into three parts, the total being represented by PA +PB +PC . As P moves closer to A' the length PA increases and when P coincides with A' we have PB = PC = 0. We can label the vertices of the triangle to represent the three parts so that PA gives the percentage of A', PB gives the percentage of B', and PC gives the percentage of C'. Suppose, for example, these percentages are 60%, 30% and 10% respectively. If we draw an equilateral triangle with sides of length L = 200/ <mdo:image align="bottom" alt="*" border="0" src="root.gif"></mdo:image> 3 then PA + PB + PC = 100.</td> </tr> <comment> Spacer row --&gt; &lt;tr&gt; &lt;td colspan="2" width="590"&gt;&#160;&lt;/td&gt; &lt;/tr&gt; &lt;tr&gt; &lt;td valign="top" width="300"&gt;To find the position of P draw lines parallel to the sides of the triangle as shown in the diagram such that anywhere on the 60% line the perpendicular distance between this line and the line B'C' = 60 so PA = 60, similarly the distance between the 30% line and the line C'A' is 30 units so that PB = 30, and the distance between the 10% line and the line A'B' is 10 units so that PC = 10. Because PA + PB + PC = 100 (as we have shown) these three lines intersect in a single point.&lt;/td&gt; &lt;td valign="top" width="290"&gt;&lt;img alt="Figure 2" src="/content/97/09/six6/sept_fig2.gif" /&gt;&lt;/td&gt; &lt;/tr&gt; &lt;tr&gt;&lt;!-- Spacer row </comment> <td colspan="2" width="590" style=""> </td> 2 3 0 0 0 0 0 Summer students' sandbox 2012 summer students' learning place