1973 AHSME Problems/Problem 35

Problem

In the unit circle shown in the figure, chords $PQ$ and $MN$ are parallel to the unit radius $OR$ of the circle with center at $O$. Chords $MP$, $PQ$, and $NR$ are each $s$ units long and chord $MN$ is $d$ units long.

[asy] draw(Circle((0,0),10)); draw((0,0)--(10,0)--(8.5,5.3)--(-8.5,5.3)--(-3,9.5)--(3,9.5)); dot((0,0)); dot((10,0)); dot((8.5,5.3)); dot((-8.5,5.3)); dot((-3,9.5)); dot((3,9.5)); label("1", (5,0), S); label("s", (8,2.6)); label("d", (0,4)); label("s", (-5,7)); label("s", (0,8.5)); label("O", (0,0),W); label("R", (10,0), E); label("M", (-8.5,5.3), W); label("N", (8.5,5.3), E); label("P", (-3,9.5), NW); label("Q", (3,9.5), NE);  [/asy]

Of the three equations

\[\textbf{I.}\ d-s=1, \qquad \textbf{II.}\ ds=1, \qquad \textbf{III.}\ d^2-s^2=\sqrt{5}\]

those which are necessarily true are

$\textbf{(A)}\ \textbf{I}\ \text{only} \qquad\textbf{(B)}\ \textbf{II}\ \text{only} \qquad\textbf{(C)}\ \textbf{III}\ \text{only} \qquad\textbf{(D)}\ \textbf{I}\ \text{and}\ \textbf{II}\ \text{only} \qquad\textbf{(E)}\ \textbf{I, II}\ \text{and} \textbf{ III}$

Solution 1

[asy] pair O=(0,0), R=(10,0), M=(-8.5,5.3), N=(8.5,5.3), P=(-3,9.5), Q=(3,9.5);  draw(Circle(O,10)); draw(O--R--N--M--P--Q); dot(O); dot(R); dot(N); dot(M); dot(P); dot(Q); label("1", (5,0), S); label("s", (10.3,2.6));  label("s", (-5,7)); label("s", (0,8.5)); label("O", O, S); label("R", R, E); label("M", M, W); label("N", N, E); label("P", P, NW); label("Q", Q, NE);  dot((-10,0)); draw(Q--N); label("S",(-10,0),W); draw(M--(-10,0)--O--cycle); draw(M--O--N,dotted); draw(P--R);  [/asy]

First, let $S$ be on circle $O$ so $RS$ is a diameter. In order to prove that the three statements are true (or false), we first show that $SM = QN = s,$ and then we examine each statement one by one.


Lemma 1: $SM = QN = s$
Since $OM = ON,$ by the Base Angle Theorem, $\angle OMN = \angle ONM.$ By the Alternating Interior Angle Theorem, $\angle MOS = \angle NOR,$ making $\triangle MOS \cong NOR$ by SAS Congruency. That means $SM = NR = s$ by CPCTC.


Because $PQNM$ is a cyclic quadrilateral, $\angle MPQ + \angle QNM = 180^\circ,$ so $\angle PMN = \angle QNM.$ That makes $PQNM$ an isosceles trapezoid, so $MP = QN = s.$


Lemma 2: Showing Statement I is (or isn’t) true

[asy] pair O=(0,0), R=(10,0), M=(-8.5,5.3), N=(8.5,5.3), P=(-3,9.5), Q=(3,9.5);  draw(Circle(O,10)); draw(O--R--N--M--P--Q); dot(O); dot(R); dot(N); dot(M); dot(P); dot(Q); label("1", (5,0), S); label("s", (10.3,2.6));  label("s", (-5,7)); label("s", (0,8.5)); label("O", O, S); label("R", R, E); label("M", M, W); label("N", N, E); label("P", P, NW); label("Q", Q, NE);  dot((-10,0)); draw(Q--N); label("S",(-10,0),W); draw(M--(-10,0)--O--cycle); draw(P--R); draw(M--R,dotted);  pair X=intersectionpoint(M--N, P--R); dot(X); label("X",X,S);  [/asy]

Let $X$ be the intersection of $MN$ and $PR.$ By SSS Congruency, $\triangle POQ \cong \triangle QON \cong \triangle NOR,$ so $\angle PQN = \angle QNR.$ We know that $PQNR$ is a cyclic quadrilateral, so $\angle PQN + \angle PRN = \angle QNR + \angle PRN = 180^\circ,$ so $QN \parallel PR.$ That makes $PQNX$ a parallelogram, so $XN = s.$ Thus, $MX = d-s.$


In addition, $PX = s,$ and by the Base Anlge Theorem and Vertical Angle Theorem, $\angle PMX = \angle MXP = \angle NXR = \angle NRX.$ That means by AAS Congruency, $\triangle MXP = \triangle RXN,$ so $MX = XR = d-s$.


By the Base Angle Theorem and the Alternating Interior Angle Theorem, $\angle XRM = \angle XMR = \angle MRO = \angle RMO,$ so by ASA Congruency, $\angle MXR = \angle MOR.$ Thus, $MX = OR = d-s = 1.$ Statement I is true.


Lemma 3: Showing Statement II is (or isn’t) true

[asy] pair O=(0,0), R=(10,0), M=(-8.5,5.3), N=(8.5,5.3), P=(-3,9.5), Q=(3,9.5);  draw(Circle(O,10)); draw(O--R--N--M--P--Q); dot(O); dot(R); dot(N); dot(M); dot(P); dot(Q); label("1", (5,0), S); label("s", (10.3,2.6));  label("s", (-5,7)); label("s", (0,8.5)); label("O", O, S); label("R", R, E); label("M", M, W); label("N", N, E); label("P", P, NW); label("Q", Q, NE);  dot((-10,0)); draw(Q--N); label("S",(-10,0),W); draw(M--(-10,0)--O--cycle); draw(P--R); draw(P--(-6,0),dotted); dot((-6,0)); label("Y",(-6,0),S);  pair X=intersectionpoint(M--N, P--R); dot(X); label("X",X,S);  [/asy]

From Lemma 2, we have $PX + XR = s + (d-s) = d$. Draw point $Y$ on $SR$ such that $YR = d,$ making $YO = s.$


Since $PQ \parallel YO,$ we have $\angle QPO = \angle POY.$ Additionally, $PQ = YO$ and $PO = PO,$ so by SAS Congruency, $\triangle PQY \cong \triangle OYQ.$ That means $PY = OQ = 1.$


Since $\angle PRS$ is an inscribed angle, $\angle POS = 2 \cdot \angle PRS.$ Additionally, $\triangle SOM \cong \triangle MOP$, so $MO$ bisects $\angle SOP.$ Thus, $\angle SOM = \angle PRS,$ making $\triangle PRY \sim \triangle MOS$ by SAS Similarity.


By using the similarity, we find that \begin{align*} \frac{1}{s} &= \frac{d}{1} \\ ds &= 1 \end{align*} Thus, Statement II is true.


Lemma 4: Showing Statement III is (or isn’t) true
From Lemmas 2 and 3, we have $d-s = 1$ and $ds = 1$. Squaring the first equation results in \[d^2 - 2ds + s^2 = 1.\] Adding $4ds = 4$ to both sides results in \begin{align*} d^2 + 2ds + s^2 &= 5 \\ (d+s)^2 &= 5 \end{align*} Since $d+s$ is positive, we find that $d+s = \sqrt{5},$ which confirms that Statement III is true.


In summary, all three statements are true, so the answer is $\boxed{\textbf{(E)}}.$


Solution 2

It is a well-known fact that any cyclic trapezoid has its legs equal. Therefore, $QN = s$. Now, extend $OR$ to meet the circle again at $T$. By similar reasoning, $MT = s$. Furthermore, since $\angle MOT, \angle POM, \angle POQ, \angle QON,$ and $\angle NOR$ sum to $180$ and are equal, they have measure $36$ degrees. It trivially follows that $2 \sin 18 = s \Longleftrightarrow s= 2\left(\frac{-1+\sqrt{5}}{4} \right)$. Dropping the altitude from $O$ to $MN$, we see that $d = 2\sin 54 \Longleftrightarrow d = 2\left(\frac{1+\sqrt{5}}{4} \right)$. Therefore, $d-s=1$, $ds=1$, and $d^2-s^2 = \sqrt{5}$.

See Also

1973 AHSME (ProblemsAnswer KeyResources)
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