Difference between revisions of "1973 AHSME Problems/Problem 35"
Made in 2016 (talk | contribs) (→See Also) |
Euler12345 (talk | contribs) m (→Solution 1) |
||
(3 intermediate revisions by 2 users not shown) | |||
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> | <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> | <asy> | ||
Line 72: | Line 72: | ||
<br> | <br> | ||
'''Lemma 1: <math>SM = QN = s</math>'''<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 | + | 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> | <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> | <br> | ||
Line 194: | Line 194: | ||
<br> | <br> | ||
In summary, all three statements are true, so the answer is <math>\boxed{\textbf{(E)}}.</math> | 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== | ==See Also== |
Latest revision as of 19:41, 17 April 2022
Contents
[hide]Problem
In the unit circle shown in the figure, chords and are parallel to the unit radius of the circle with center at . Chords , , and are each units long and chord is units long.
Of the three equations
those which are necessarily true are
Solution 1
First, let be on circle so is a diameter. In order to prove that the three statements are true (or false), we first show that and then we examine each statement one by one.
Lemma 1:
Since by the Base Angle Theorem, By the Alternate Interior Angles Theorem, making by SAS Congruency. That means by CPCTC.
Because is a cyclic quadrilateral, , but we are given that is parallel to , so . Therefore, That makes an isosceles trapezoid, so
Lemma 2: Showing Statement I is (or isn’t) true
Let be the intersection of and By SSS Congruency, so We know that is a cyclic quadrilateral, so so That makes a parallelogram, so Thus,
In addition, and by the Base Anlge Theorem and Vertical Angle Theorem, That means by AAS Congruency, so .
By the Base Angle Theorem and the Alternating Interior Angle Theorem, so by ASA Congruency, Thus, Statement I is true.
Lemma 3: Showing Statement II is (or isn’t) true
From Lemma 2, we have . Draw point on such that making
Since we have Additionally, and so by SAS Congruency, That means
Since is an inscribed angle, Additionally, , so bisects Thus, making by SAS Similarity.
By using the similarity, we find that
Thus, Statement II is true.
Lemma 4: Showing Statement III is (or isn’t) true
From Lemmas 2 and 3, we have and . Squaring the first equation results in
Adding to both sides results in
Since is positive, we find that which confirms that Statement III is true.
In summary, all three statements are true, so the answer is
Solution 2
It is a well-known fact that any cyclic trapezoid has its legs equal. Therefore, . Now, extend to meet the circle again at . By similar reasoning, . Furthermore, since and sum to and are equal, they have measure degrees. It trivially follows that . Dropping the altitude from to , we see that . Therefore, , , and .
See Also
1973 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 34 |
Followed by Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |