Difference between revisions of "1973 AHSME Problems/Problem 35"

Latest revision as of 19:41, 17 April 2022

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 Alternate Interior Angles Theorem, $\angle MOS = \angle NMO = \angle MNO = \angle NOR,$ making $\triangle MOS \cong NOR$ by SAS Congruency. That means $SM = NR = s$ by CPCTC.

Because $PQNM$ is a cyclic quadrilateral, $\angle MPQ + \angle QNM = 180^\circ$ , but we are given that $PQ$ is parallel to $MN$ , so $\angle MPQ + \angle PMN = 180^\circ$ . Therefore, $\angle PMN = \angle QNM.$ 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 $d^2-s^2 = \sqrt{5}$ .

@@ Line 34: / Line 34: @@
 <math>\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}</math>
-==Solution==
+==Solution 1==
 <asy>
@@ Line 72: / Line 72: @@
 <br>
 '''Lemma 1: <math>SM = QN = s</math>'''<br>
-Since <math>OM = ON,</math> by the Base Angle Theorem, <math>\angle OMN = \angle ONM.</math>  By the Alternating Interior Angle Theorem, <math>\angle MOS = \angle NOR,</math> making <math>\triangle MOS \cong NOR</math> by SAS Congruency.  That means <math>SM = NR = s</math> by CPCTC.
+Since <math>OM = ON,</math> by the Base Angle Theorem, <math>\angle OMN = \angle ONM.</math>  By the Alternate Interior Angles Theorem, <math>\angle MOS = \angle NMO = \angle MNO = \angle NOR,</math> making <math>\triangle MOS \cong NOR</math> by SAS Congruency.  That means <math>SM = NR = s</math> by CPCTC.
 <br>
-Because <math>PQNM</math> is a [[cyclic quadrilateral]], <math>\angle MPQ + \angle QNM = 180^\circ,</math> so <math>\angle PMN = \angle QNM.</math>  That makes <math>PQNM</math> an [[isosceles trapezoid]], so <math>MP = QN = s.</math>
+Because <math>PQNM</math> is a [[cyclic quadrilateral]], <math>\angle MPQ + \angle QNM = 180^\circ</math>, but we are given that <math>PQ</math> is parallel to <math>MN</math>, so <math>\angle MPQ + \angle PMN = 180^\circ</math>. Therefore, <math>\angle PMN = \angle QNM.</math>  That makes <math>PQNM</math> an [[isosceles trapezoid]], so <math>MP = QN = s.</math>
 <br>
@@ Line 194: / Line 194: @@
 <br>
 In summary, all three statements are true, so the answer is <math>\boxed{\textbf{(E)}}.</math>
+==Solution 2==
+It is a well-known fact that any cyclic trapezoid has its legs equal. Therefore, <math>QN = s</math>. Now, extend <math>OR</math> to meet the circle again at <math>T</math>. By similar reasoning, <math>MT = s</math>. Furthermore, since <math>\angle MOT, \angle POM, \angle POQ, \angle QON,</math> and <math>\angle NOR</math> sum to <math>180</math> and are equal, they have measure <math>36</math> degrees. It trivially follows that <math>2 \sin 18 = s \Longleftrightarrow s= 2\left(\frac{-1+\sqrt{5}}{4} \right)</math>. Dropping the altitude from <math>O</math> to <math>MN</math>, we see that <math>d = 2\sin 54 \Longleftrightarrow d = 2\left(\frac{1+\sqrt{5}}{4} \right)</math>. Therefore, <math>d-s=1</math>, <math>ds=1</math>, and <math>d^2-s^2 = \sqrt{5}</math>.
 ==See Also==
-{{AHSME 35p box|year=1973|num-b=34|after=Last Problem}}
+{{AHSME 30p box|year=1973|num-b=34|after=Last Problem}}
 [[Category:Intermediate Geometry Problems]]

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Difference between revisions of "1973 AHSME Problems/Problem 35"

Latest revision as of 19:41, 17 April 2022

Contents

Problem

Solution 1

Solution 2

See Also