CEMC Galois, 2018

Geometry Difficulty 3.1 AMC 10/12 Prove it Canada

Here are two facts about circles:

If points $A$ , $B$ , $C$ lie on a circle so that $\angle ABC=90^{\circ}$ , then $AC$ is a diameter of the circle. This means that in Figure 1, $AC$ is a diameter of the circle.
If points $D$ , $E$ , $F$ lie on a circle so that $EF$ is a diameter, then $\angle EDF=90^{\circ}$ . This means that in Figure 2, $\angle EDF=90^{\circ}$ .

In Figure 1 above, $AB=8$ and $BC=15$ . What is the length of diameter $AC$ ?
In Figure 2 above, $DE=24$ and the radius of the circle is 13. What is the length of $DF$ ?
In Figure 3, points $P$ , $Q$ , $R$ , and $S$ are on a circle with centre $O$ . Also, $SQ$ is a diameter of the circle and $O$ is joined to $R$ . If $SP=PQ$ and $\angle RQP=80^{\circ}$ , determine the measure of $\angle ROQ$ and the measure of $\angle RSQ$ .

Solution

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

Using the Pythagorean Theorem, we get $AC^2=AB^2+BC^2$ or $AC^2=8^2+15^2$ , and so $AC=\sqrt{64+225}=\sqrt{289}=17$ (since $\text{AC gt;0}$ ).
In Figure 2, $EF$ is a diameter and so its length is twice the radius or 26.

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

Using the Pythagorean Theorem, we get $DF^2=EF^2-DE^2$ or $DF^2=26^2-24^2$ , and so $DF=\sqrt{676-576}=\sqrt{100}=10$ (since $\text{DF gt;0}$ ).
Since $SQ$ is a diameter, then $\angle SPQ=\angle SRQ=90^{\circ}$ .

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

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

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

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

$\angle ROQ= 180^{\circ}-2\times35^{\circ}=110^{\circ}$ .

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

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