Difference between revisions of "2017 UNM-PNM Statewide High School Mathematics Contest II Problems/Problem 8"
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== Solution== | == Solution== | ||
+ | Method 1: | ||
+ | We can compute the area in two ways: <math>\triangle{ABC} = \frac{1}{2}\cdot 5\cdot 12 = 30</math> or <math>\triangle{ABC}=\triangle{ABX} +\triangle{CBX} \frac{1}{2}\cdot{r}\cdot{13}+\frac{1}{2}\cdot{r}\cdot{5} = 9r</math>. Setting the two areas equal we obtain <math>r = \frac{10}{3}</math>. | ||
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+ | Method 2: | ||
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+ | Point <math>X</math> lies at cartesian coordinate <math>(0,r)</math>. | ||
+ | The line AB has formula <math>y=\frac{12}{5}\cdot(5-x)</math>. | ||
+ | The vector <math>\vec{XY}</math> has coordinates <math>r(12/13,5/13)</math> since it has length <math>r</math> in the unit direction <math>(12,5)/13</math> which is orthogonal to the line AB. | ||
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+ | So the point Y has coordinates <math>X + \vec{XY}=(0,r)+r(12/13,5/13)</math> and lies on the line <math>y=\frac{12}{5}\cdot(5-x)</math>. | ||
+ | Substituting for these equations gives <math>r = \frac{10}{3}</math>. | ||
== See also == | == See also == |
Revision as of 04:34, 19 January 2019
Problem
Let be a right triangle with right angle at . Suppose and and is the diameter of a semicircle, where lies on and the semicircle is tangent to side . Find the radius of the semicircle.
Solution
Method 1:
We can compute the area in two ways: or . Setting the two areas equal we obtain .
Method 2:
Point lies at cartesian coordinate . The line AB has formula . The vector has coordinates since it has length in the unit direction which is orthogonal to the line AB.
So the point Y has coordinates and lies on the line . Substituting for these equations gives .
See also
2017 UNM-PNM Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNM-PNM Problems and Solutions |