Difference between revisions of "1978 AHSME Problems/Problem 26"
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We know that triangle <math>RCQ</math> is similar to triangle <math>ABC</math>. We draw a line to point <math>D</math> on hypotenuse <math>AB</math> such that <math>\angle QDR</math> is <math>90 ^\circ</math> and that <math>RDQC</math> is a rectangle. Since triangle <math>RCQ</math> is similar to triangle <math>ABC</math>, let <math>RC</math> be <math>4x</math> and <math>RD/CQ</math> be <math>3x</math>. Now we have line segment <math>AQ</math> = <math>8-3x</math>, and line segment <math>RB</math> = <math>6-4x</math>. Since <math>BD + DA = AB</math>, we use simple algebra and Pythagorean Theorem to get <math>\sqrt {(3x)^2 + (6-4x)^2}</math> + <math>\sqrt {(4x)^2 + (8-3x)^2}</math> = <math>10</math>. Expanding and simplifying gives us <math>\sqrt {25x^2-48x+36}</math> + <math>\sqrt {25x^2-48x+64}</math> = <math>10</math>. | We know that triangle <math>RCQ</math> is similar to triangle <math>ABC</math>. We draw a line to point <math>D</math> on hypotenuse <math>AB</math> such that <math>\angle QDR</math> is <math>90 ^\circ</math> and that <math>RDQC</math> is a rectangle. Since triangle <math>RCQ</math> is similar to triangle <math>ABC</math>, let <math>RC</math> be <math>4x</math> and <math>RD/CQ</math> be <math>3x</math>. Now we have line segment <math>AQ</math> = <math>8-3x</math>, and line segment <math>RB</math> = <math>6-4x</math>. Since <math>BD + DA = AB</math>, we use simple algebra and Pythagorean Theorem to get <math>\sqrt {(3x)^2 + (6-4x)^2}</math> + <math>\sqrt {(4x)^2 + (8-3x)^2}</math> = <math>10</math>. Expanding and simplifying gives us <math>\sqrt {25x^2-48x+36}</math> + <math>\sqrt {25x^2-48x+64}</math> = <math>10</math>. | ||
− | Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by <math>\sqrt {25x^2-48x+36}</math>. Now, we can square both sides and simplify to get <math>0 = 72 - 20 \sqrt{25x^2-48x+36}</math>. Dividing both sides by <math>4</math>, we get <math>18 - 5 \sqrt {25x^2-48x+36}</math> = <math>0</math>. We then add <math>5 \sqrt {25x^2-48x+36}</math> to both sides to get <math>18 = 5 \sqrt {25x^2-48x+36}</math>. Since this is very messy, let <math>25x^2 - 48x = y</math>. | + | Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by <math>\sqrt {25x^2-48x+36}</math>. Now, we can square both sides and simplify to get <math>0 = 72 - 20 \sqrt{25x^2-48x+36}</math>. Dividing both sides by <math>4</math>, we get <math>18 - 5 \sqrt {25x^2-48x+36}</math> = <math>0</math>. We then add <math>5 \sqrt {25x^2-48x+36}</math> to both sides to get <math>18 = 5 \sqrt {25x^2-48x+36}</math>. Since this is very messy, let <math>25x^2 - 48x = y</math>. Squaring both sides, we get <math>324 = 25y + 900, 25y = -576</math>. Solving for <math>y</math>, we have <math>y = -23.04</math>. Plugging in <math>y</math> as <math>25x^2-48x</math>, we have <math>25x^2-48x+23.04 = 0</math>. Using the quadratic equation, we get <math>\frac {48+0}{50}</math>. Therefore, <math>x = \frac {48}{50}</math>. |
Remember that our desired answer is the hypotenuse of the triangle <math>3x - 4x - 5x</math>. Since <math>5x</math> is the hypotenuse, our answer is <math>\boxed {(B)4.8}</math> | Remember that our desired answer is the hypotenuse of the triangle <math>3x - 4x - 5x</math>. Since <math>5x</math> is the hypotenuse, our answer is <math>\boxed {(B)4.8}</math> |
Latest revision as of 19:28, 6 November 2021
Problem
In and
. Circle
is the circle with smallest radius which passes through
and is tangent to
. Let
and
be the points of intersection, distinct from
, of circle
with sides
and
, respectively. The length of segment
is
Solution
We know that triangle is similar to triangle
. We draw a line to point
on hypotenuse
such that
is
and that
is a rectangle. Since triangle
is similar to triangle
, let
be
and
be
. Now we have line segment
=
, and line segment
=
. Since
, we use simple algebra and Pythagorean Theorem to get
+
=
. Expanding and simplifying gives us
+
=
.
Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by . Now, we can square both sides and simplify to get
. Dividing both sides by
, we get
=
. We then add
to both sides to get
. Since this is very messy, let
. Squaring both sides, we get
. Solving for
, we have
. Plugging in
as
, we have
. Using the quadratic equation, we get
. Therefore,
.
Remember that our desired answer is the hypotenuse of the triangle . Since
is the hypotenuse, our answer is
~Arcticturn
See Also
1978 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 25 |
Followed by Problem 27 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.