Difference between revisions of "1973 AHSME Problems/Problem 35"
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− | Because <math>PQNM</math> is a [[cyclic quadrilateral]], <math>\angle MPQ + \angle QNM = 180^\circ</math>, but we are given that PQ is parallel to MN, 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> | + | 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> |
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Latest revision as of 20:41, 17 April 2022
Contents
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 |