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Olympiad workbook
Exam
Olympiad workbook
Subject
IOQM
Chapter
Circle

Olympiad workbook · IOQM · Circle

Consider an isosceles triangle A B C with sides B C=30, C A=A B=20...

Consider an isosceles triangle ABCA B C with sides BC=30,CA=AB=20B C=30, C A=A B=20. Let DD be the foot of the perpendicular from AA to BCB C, and let MM be the midpoint of ADA D. Let PQP Q be a chord of the circumcircle of triangle ABCA B C, such that MM lies on PQP Q is parallel to BCB C. The length of PQP Q is:Consider an isosceles triangle A B C with sides B C=30, C A=A B=20 . Let D be the foot of the perpendicular from A to B

Correct answer

25

Step-by-step solution

Eq. of PQ:y=572P Q: y=\frac{5 \sqrt{7}}{2} Consider an isosceles triangle A B C with sides B C=30, C A=A B=20 . Let D be the foot of the perpendicular from A to B
Perpendicular bisector of ACA C : (y−572)=−1(57−00−15)(x−152)⇒(y−572)=1557(x−152) Centre ≡ intersection of perpendicular bisector ≡(0,−57)≡(0,−577)⇒ Eq n  of circumcircle \begin{aligned} & \left(y-\frac{5 \sqrt{7}}{2}\right)=\frac{-1}{\left(\frac{5 \sqrt{7}-0}{0-15}\right)}\left(x-\frac{15}{2}\right) \\ & \Rightarrow \quad\left(y-\frac{5 \sqrt{7}}{2}\right)=\frac{15}{5 \sqrt{7}}\left(x-\frac{15}{2}\right) \\ & \text { Centre } \equiv \text { intersection of perpendicular bisector } \\ & \equiv\left(0, \frac{-5}{\sqrt{7}}\right) \equiv\left(0, \frac{-5 \sqrt{7}}{7}\right) \\ & \Rightarrow \quad \text { Eq }{ }^{\text {n }} \text { of circumcircle } \end{aligned} (x−0)2+(y+577)2=(57+577)2x2+(y+57)2=25×7×6449=25×647\begin{aligned} & (x-0)^2+\left(y+\frac{5 \sqrt{7}}{7}\right)^2=\left(5 \sqrt{7}+\frac{5 \sqrt{7}}{7}\right)^2 \\ & x^2+\left(y+\frac{5}{\sqrt{7}}\right)^2=25 \times 7 \times \frac{64}{49}=\frac{25 \times 64}{7} \end{aligned} Intersection with PQ:y=572P Q: y=\frac{5 \sqrt{7}}{2} x2=25×647−(572+57)2=25×647−14×7(452)x2=25×64×4−45228=5222(64×4−817)=5222×25⇒x=±252⇒ distance =∣252−(−252)∣=25\begin{aligned} & x^2=\frac{25 \times 64}{7}-\left(\frac{5 \sqrt{7}}{2}+\frac{5}{\sqrt{7}}\right)^2=\frac{25 \times 64}{7}-\frac{1}{4 \times 7}\left(45^2\right) \\ & x^2=\frac{25 \times 64 \times 4-45^2}{28}=\frac{5^2}{2^2}\left(\frac{64 \times 4-81}{7}\right)=\frac{5^2}{2^2} \times 25 \\ & \Rightarrow \quad x= \pm \frac{25}{2} \Rightarrow \text { distance }=\left|\frac{25}{2}-\left(-\frac{25}{2}\right)\right|=25 \end{aligned}
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