Maths Olympiad Prep

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Geometry Difficulty 2.6 Junior Find the answer Canada

In the diagram, AA, BB, DD, FF, and GG lie on a vertical line, BCD\triangle BCD is right-angled at CC, and DEF\triangle DEF is right-angled at EE. Also, ABC=x°\angle ABC = x\degree, CDE=80°\angle CDE = 80\degree, and EFG=y°\angle EFG = y\degree.Figure 0What is the value of x+yx+y?

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Solution

Since ABD=180°\angle ABD = 180\degree and ABC=x°\angle ABC = x\degree, then CBD=180°x°\angle CBD = 180\degree - x\degree.

Since the measures of the angles in BCD\triangle BCD add to 180°180\degree, then BDC=180°(180°x°)90°=x°90°\angle BDC = 180\degree - (180\degree - x\degree) - 90\degree = x\degree - 90\degree Similarly, GFD=180°\angle GFD = 180\degree and FDE=y°90°\angle FDE = y\degree - 90\degree. Finally, BDF=180°\angle BDF = 180\degree and so BDC+CDE+FDE=180°(x°90°)+80°+(y°90°)=180°x+y100=180\begin{align*} \angle BDC + \angle CDE + \angle FDE & = 180\degree \\ (x\degree - 90\degree) + 80\degree + (y\degree - 90\degree) & = 180\degree \\ x + y - 100 & = 180\end{align*} and so x+y=280x + y = 280.

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Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.