Difference between revisions of "1968 AHSME Problems/Problem 7"
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\text{(B) } \frac{9}{2}\quad | \text{(B) } \frac{9}{2}\quad | ||
\text{(C) } 6\quad | \text{(C) } 6\quad | ||
| − | \text{(D) } | + | \text{(D) } 9\quad |
\text{(E) } \text{undetermined}</math> | \text{(E) } \text{undetermined}</math> | ||
== Solution == | == Solution == | ||
| − | <math>\fbox{}</math> | + | The fact that <math>OQ=3</math> only tells us that <math>CQ=9</math>. There are infinitely many triangles with a median which has length 9, so we can make no statement about the length of the median <math>\overline{AP}</math> or the segment <math>\overline{OP}</math>. Thus, <math>OP</math> is <math>\fbox{(E) undetermined}</math>. |
== See also == | == See also == | ||
| − | {{AHSME box|year=1968|num-b=6|num-a=8}} | + | {{AHSME 35p box|year=1968|num-b=6|num-a=8}} |
[[Category: Introductory Geometry Problems]] | [[Category: Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 19:13, 17 July 2024
Problem
Let 
be the intersection point of medians 
and 
of triangle 
if 
is 3 inches, then 
, in inches, is:
Solution
The fact that 
only tells us that 
. There are infinitely many triangles with a median which has length 9, so we can make no statement about the length of the median 
or the segment 
. Thus, is 
.
See also
| 1968 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
