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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Circle

Olympiad workbook · IOQM · Circle

Let A B be a diameter of a circle and let C be a point on the...

Let ABA B be a diameter of a circle and let CC be a point on the segment ABA B such that AC:CB=6:7A C: C B=6: 7. Let DD be a point on the circle such that DCD C is perpendicular to ABA B. Let DED E be the diameter through DD. If [XYZ][X Y Z] denotes the area of the triangle XYZX Y Z. Find [ABD]/[CDE][A B D] /[C D E] to the nearest integer.Let A B be a diameter of a circle and let C be a point on the segment A B such that A C: C B=6: 7 . Let D be a point on

Correct answer

13

Step-by-step solution

Let A B be a diameter of a circle and let C be a point on the segment A B such that A C: C B=6: 7 . Let D be a point on
CD=r2−r2132=r1312×14tan⁡α=112×14⇒sin⁡α=112×14+1CPCD=sin⁡α=CPr1312×14=112×14+1CP=r12×141312×14+1∣ABD∣∣CDE∣=12(2r)×r1312×1412(2r)×r1312×1412×14+1=12×14+1=13\begin{aligned} & C D=\sqrt{r^2-\frac{r^2}{13^2}}=\frac{r}{13} \sqrt{12 \times 14} \\ & \tan \alpha=\frac{1}{\sqrt{12 \times 14}} \Rightarrow \sin \alpha=\frac{1}{\sqrt{12 \times 14+1}} \\ & \frac{C P}{C D}=\sin \alpha=\frac{C P}{\frac{r}{13} \sqrt{12 \times 14}}=\frac{1}{\sqrt{12 \times 14+1}} \\ & C P=\frac{r \sqrt{12 \times 14}}{13 \sqrt{12 \times 14+1}} \\ & \frac{|A B D|}{|C D E|}=\frac{\frac{1}{2}(2 r) \times \frac{r}{13} \sqrt{12 \times 14}}{\frac{1}{2}(2 r) \times \frac{r}{13} \frac{\sqrt{12 \times 14}}{\sqrt{12 \times 14+1}}}=\sqrt{12 \times 14+1}=13\end{aligned}
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