hi guys ive be struggling on this problem for a couple of days, so please help if you can
Problem:Two altitudes of a triangle have lengths 12 and 14. What is the longest possible length of the third altitude, if it is a positive integer?
Thanks in advance and merry christmas: happy new year!
I'm not sure there is one. Consider an isosceles triangle, where the two shorter altitudes are the same, say h.
If the distance from the base of the altitude to a base vertex is c, then the base angle θ is such that
tanθ = h/c
If the remaining altitude is k, then we also have
tanθ = k/(√(h^2+c^2)/2) = h/c
k = h√(h^2+c^2)/2c
as c gets smaller and smaller, k gets closer and closer to h^2/2c, which can grow without limit.
Granted, we don't have an isosceles triangle, but the two altitudes are almost equal, and I think the same argument could be made.
I tried a different approach, but I don't like my last sequence of steps.
What is your opinion ?
http://www.jiskha.com/display.cgi?id=1357058982
I even tried playing with the triangle here ...
http://www.mathopenref.com/triangleorthocenter.html
My argument above fails because we don't have an isosceles triangle. In the limit, with an isosceles triangle, we end up with an infinitely tall "triangle" with parallel sides h units apart. However, because we have diagonals 12 and 14, that is not possible.
More thought required.
Two altitudes of a triangle have lengths 12 and 14. What is the longest possible length of the third altitude, if it is a positive integer?
To find the longest possible length of the third altitude of the triangle, we will first need to understand the properties of altitudes in a triangle.
An altitude of a triangle is a line segment drawn from a vertex of the triangle to the opposite side, forming a right angle. The length of an altitude determines the height of the triangle from that particular vertex.
In this problem, we are given two altitudes with lengths 12 and 14. Let's assume that these altitudes correspond to the vertices opposite to them (for example, the altitude of length 12 corresponds to the side opposite to vertex A).
To find the longest possible length of the third altitude, we need to consider the relationship between the lengths of the sides of a triangle and its altitudes. The area of a triangle is given by the formula:
Area = (1/2) * base * height
Where the base is the length of one side of the triangle and the height is the length of the corresponding altitude. Since we are looking for the longest possible length of the third altitude, we need to maximize the area of the triangle.
To maximize the area of a triangle, we need to maximize both the lengths of the base and the corresponding altitude. In our case, we already have two altitudes with lengths 12 and 14.
To maximize the length of the third altitude, we can consider the fact that a longer base will give us a larger area. The base needs to be at least as long as the sum of the lengths of the other two sides of the triangle (according to the Triangle Inequality Theorem). Therefore, the third side (base) needs to be at least 12 + 14 = 26.
Now, to find the longest possible length of the third altitude, we need to consider the possible lengths of the base. Since the base needs to be a positive integer, we can start from 26 and work our way up. We will calculate the corresponding area of the triangle for each possible base length.
Let's start with a base length of 26:
Area = (1/2) * 26 * 12 = 156
Area = (1/2) * 26 * 14 = 182
As we increase the length of the base, the area of the triangle will increase. However, we are looking for the largest area that corresponds to a positive integer as the third altitude.
By trying out different values for the base length, we can find that the longest possible length of the third altitude is achieved when the base length is 68.
Area = (1/2) * 68 * 12 = 408 (a positive integer)
Therefore, the longest possible length of the third altitude is 68.
Merry Christmas and Happy New Year to you too! I hope this explanation helps.