Maths Olympiad Prep

Library / /11 of 30

, 2018

Geometry Difficulty 3.1 AMC 10/12 Prove it Canada

Here are two facts about circles:

If points AA, BB, CC lie on a circle so that ABC=90\angle ABC=90^{\circ}, then ACAC is a diameter of the circle. This means that in Figure 1, ACAC is a diameter of the circle.
If points DD, EE, FF lie on a circle so that EFEF is a diameter, then EDF=90\angle EDF=90^{\circ}. This means that in Figure 2, EDF=90\angle EDF=90^{\circ}.

In Figure 1 above, AB=8AB=8 and BC=15BC=15. What is the length of diameter ACAC?
In Figure 2 above, DE=24DE=24 and the radius of the circle is 13. What is the length of DFDF?
In Figure 3, points PP, QQ, RR, and SS are on a circle with centre OO. Also, SQSQ is a diameter of the circle and OO is joined to RR. If SP=PQSP=PQ and RQP=80\angle RQP=80^{\circ}, determine the measure of ROQ\angle ROQ and the measure of RSQ\angle RSQ.

Solution

In ABC\triangle ABC, ABC=90\angle ABC=90^{\circ}.

Using the Pythagorean Theorem, we get AC2=AB2+BC2AC^2=AB^2+BC^2 or AC2=82+152AC^2=8^2+15^2, and so AC=64+225=289=17AC=\sqrt{64+225}=\sqrt{289}=17 (since AC>0AC>0).
In Figure 2, EFEF is a diameter and so its length is twice the radius or 26.

From the second fact, we know that EDF=90\angle EDF=90^{\circ}.

Using the Pythagorean Theorem, we get DF2=EF2DE2DF^2=EF^2-DE^2 or DF2=262242DF^2=26^2-24^2, and so DF=676576=100=10DF=\sqrt{676-576}=\sqrt{100}=10 (since DF>0DF>0).
Since SQSQ is a diameter, then SPQ=SRQ=90\angle SPQ=\angle SRQ=90^{\circ}.

In SPQ\triangle SPQ, SP=PQSP=PQ which means that SPQ\triangle SPQ is isosceles and so

PQS=PSQ=180902=45\angle PQS=\angle PSQ=\dfrac{180^{\circ}-90^{\circ}}{2}=45^{\circ}.

Since RQP=80\angle RQP=80^{\circ}, then RQO=RQPPQS=8045=35\angle RQO=\angle RQP-\angle PQS=80^{\circ}-45^{\circ}=35^{\circ}.

In ROQ\triangle ROQ, OR=OQOR=OQ (both are radii) and so QRO=RQO=35\angle QRO=\angle RQO=35^{\circ} and

ROQ=1802×35=110\angle ROQ= 180^{\circ}-2\times35^{\circ}=110^{\circ}.

In SRQ\triangle SRQ, we get RSQ=180SRQRQS=1809035=55\angle RSQ=180^{\circ}-\angle SRQ-\angle RQS=180^{\circ}-90^{\circ}-35^{\circ}=55^{\circ}.

Want a route through all this instead of an archive? The track puts 2,444 problems in a working order, from Junior Challenge level to the IMO shortlist.

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