Art of Problem Solving

2008 UNCO Math Contest II Problems/Problem 4: Difference between revisions

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== Problem ==
== Problem ==


In the figure there are <math>8</math> line segments drawn from
vertex <math>A</math> to the base <math>BC</math> (not counting the segments <math>AB</math> or <math>AC</math>).


5. The sum of <math>400, 3, 500, 800</math> and <math>305</math> is <math>2008</math> and the product of these five numbers is
<asy>
<math>146400000000 = 1464 \times 10^8.</math>
for (int x=0;x<11;++x){
draw((5,15)--(x,0),dot);
}
draw((0,0)--(10,0),black);
draw((10/6,5)--(10-10/6,5),black);
draw((20/6,10)--(10-20/6,10),black);
MP("A",(5,15),N);MP("B",(0,0),W);MP("C",(10,0),E);
</asy>


(a) Determine the largest number which is the product of positive integers whose sum is <math>2008</math>.
(a) Determine the total number of triangles of all sizes.


(b) Determine the largest number which is the product of positive integers whose sum is <math>n</math>.
(b) How many triangles are there if there are <math>n</math> lines
drawn from <math>A</math> to <math>n</math> interior points on <math>BC</math>?




== Solution ==
== Solution ==
 
(a) <math>3\binom{10}{2}</math> (b) <math>3\binom{n+2}{2}=\frac{3(n+1)(n+2)}{2}</math>


== See Also ==
== See Also ==
{{UNCO Math Contest box|n=II|year=2008|num-b=3|num-a=5}}
{{UNCO Math Contest box|n=II|year=2008|num-b=3|num-a=5}}


[[Category:Intermediate Algebra Problems]]
[[Category:Intermediate Geometry Problems]]

Latest revision as of 01:01, 13 January 2019

Problem

In the figure there are $8$ line segments drawn from vertex $A$ to the base $BC$ (not counting the segments $AB$ or $AC$).

[asy] for (int x=0;x<11;++x){ draw((5,15)--(x,0),dot); } draw((0,0)--(10,0),black); draw((10/6,5)--(10-10/6,5),black); draw((20/6,10)--(10-20/6,10),black); MP("A",(5,15),N);MP("B",(0,0),W);MP("C",(10,0),E); [/asy]

(a) Determine the total number of triangles of all sizes.

(b) How many triangles are there if there are $n$ lines drawn from $A$ to $n$ interior points on $BC$?


Solution

(a) $3\binom{10}{2}$ (b) $3\binom{n+2}{2}=\frac{3(n+1)(n+2)}{2}$

See Also

2008 UNCO Math Contest II (ProblemsAnswer KeyResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6 7 8 9 10
All UNCO Math Contest Problems and Solutions