Difference between revisions of "1964 AHSME Problems/Problem 13"
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\textbf{(E)}\ 3:4 </math> | \textbf{(E)}\ 3:4 </math> | ||
| − | == Solution == | + | == Solution 1== |
Label our triangle <math>ABC</math> where that <math>AB=17</math>, <math>AC=13</math>, and <math>BC=8</math>. Let <math>L</math>, <math>J</math>, and <math>K</math> be the tangency points of <math>BC</math>, <math>AC</math>, and <math>AB</math> respectively. Let <math>AK=x</math>, which implies <math>AJ=x</math>. Thus <math>KB=BL=17-x</math> and <math>JC=LC=13-x</math>. | Label our triangle <math>ABC</math> where that <math>AB=17</math>, <math>AC=13</math>, and <math>BC=8</math>. Let <math>L</math>, <math>J</math>, and <math>K</math> be the tangency points of <math>BC</math>, <math>AC</math>, and <math>AB</math> respectively. Let <math>AK=x</math>, which implies <math>AJ=x</math>. Thus <math>KB=BL=17-x</math> and <math>JC=LC=13-x</math>. | ||
Since <math>BL+LC=(17-x)+(13-x)=8</math>, <math>x=11</math>. Thus <math>r:s=CL:BL=13-x:17-x=2:6=1:3</math>, hence our answer is <math>\fbox{A}</math>. | Since <math>BL+LC=(17-x)+(13-x)=8</math>, <math>x=11</math>. Thus <math>r:s=CL:BL=13-x:17-x=2:6=1:3</math>, hence our answer is <math>\fbox{A}</math>. | ||
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| + | == Solution 2== | ||
| + | |||
| + | Let the three sides of <math>8, 13, 17</math> be split into two lengths each, for a total of six segments. Because the vertices of the triangle form two congruent external tangents to the triangle each, there are really only three unique lengths, and two of these three lengths each make up a side. Thus, we can make a system of three equations in there variables: <math>a+b = 8, b+c=13, c+a=17</math>. | ||
| + | |||
| + | Adding all three equations and dividing by <math>2</math> gives <math>a+b+c = 19</math>. Subtracting each equation from that gives <math>(a, b, c) = (6,2,11)</math> | ||
| + | |||
| + | Thus, <math>a+b</math> makes up the shortest side of <math>8</math>. The ratio is <math>2:6</math>, or <math>1:3</math>, which is answer <math>\fbox{A}</math>. | ||
| + | |||
| + | ==See Also== | ||
| + | {{AHSME 40p box|year=1964|num-b=11|num-a=13}} | ||
| + | |||
| + | [[Category:Introductory Geometry Problems]] | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 17:25, 23 July 2019
Contents
Problem 13
A circle is inscribed in a triangle with side lengths 
, and 
. Let the segments of the side of length 
,
made by a point of tangency, be 
and 
, with 
. What is the ratio 
?
Solution 1
Label our triangle 
where that 
, 
, and 
. Let 
, 
, and 
be the tangency points of 
, 
, and 
respectively. Let 
, which implies 
. Thus 
and 
.
Since 
, 
. Thus 
, hence our answer is 
.
Solution 2
Let the three sides of 
be split into two lengths each, for a total of six segments. Because the vertices of the triangle form two congruent external tangents to the triangle each, there are really only three unique lengths, and two of these three lengths each make up a side. Thus, we can make a system of three equations in there variables: 
.
Adding all three equations and dividing by 
gives 
. Subtracting each equation from that gives 
Thus, 
makes up the shortest side of . The ratio is 
, or 
, which is answer .
See Also
| 1964 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
