Triangles with Proportional Sides

For each triangle $ABC$ in the set $S$ , define $N_{ABC} = \frac {a}{b} = \frac {b} {c} = \sqrt{M_{ABC}}$ . From the equilateral triangle, we have $N = 1$ , so $M=N^2=1$ and thus $C\geq 1$ . Let us look at triangles in the set $S$ which satisfy $M_{ABC} \geq 1$ , which is equivalent to $N_{ABC}\geq 1$ .

From the question, we have $b = cN$ and $a = bN = cN^2$ . The side lengths must satisfy the triangle inequality , so we have $N^2 + N> 1$ , $N^2 + 1 > N$ and $N + 1 > N^2$ . The first 2 are trivially satisfied as $N^2 \geq N \geq 1$ . From the third inequality and using the quadratic formula, we have $0 > N^2 - N - 1 \Rightarrow N < \frac {1 + \sqrt{5} } {2}$ . Thus, $2M = 2 N^2 < 3 + \sqrt{5}$ , which suggests that $2C= 3 + \sqrt{5}$ .

To show that $2C=3 + \sqrt{5}$ is the smallest real number satisfying the inequality, for each real number $r$ satisfying $1 \leq r < \frac {1+\sqrt{5} } {2}$ , there is a triangle with side lengths $1, r, r^2$ , showing that $C \geq r^2$ . Hence, we have $2C \geq 3 + \sqrt{5}$ .

Thus, $2C = 3 + \sqrt{5}$ , which gives $a+b=3+5=8$ .

So the LAST problem is to prove that the minimum of $R$ is $\frac{3+\sqrt{5}}{2}$ . Suppose that there exists triangle ABC with $M > \frac{3+\sqrt{5}}{2}$ . I want to make the solution easier, so I convert the triangle to a similar one with side $a = 1$ . Then $\frac{1}{c}\ > \frac{3+\sqrt{5}}{2}$ , so $0 < c < \frac{3-\sqrt{5}}{2}$ . By triangle inequality, $b + c > a$ , because $a = 1$ and $c < \frac{3-\sqrt{5}}{2}$ , we have $b > \frac{\sqrt{5}-1}{2}$ . $\frac{a}{b} < \frac{1}{b} < \frac{1+\sqrt{5}}{2}$ . $\frac{b}{c} > \frac{1+\sqrt{5}}{2}$ . This is a contradiction. So there is no triangles ABC with $M > \frac{3+\sqrt{5}}{2}$ . Conclusion: The minimum of $R$ is $\frac{3+\sqrt{5}}{2}$ .

We are trying to ,in a way, find the maximum value of $\frac {a}{c}$ , so let $a \geq c$ . So now we have $b= \sqrt {ac}$ . By triangle inequality, we have $\sqrt {ac} + c > a$ , $\sqrt {ac}> a-c$ , and since both LHS and RHS are positive, $ac> a^{2}+b^{2}-2ac$ , $a^2+c^2-3ac<0$ , $(\frac {a}{c})^{2}-3(\frac {a}{c})+1<0$ , $x^{2}-3x+1<0$ , where $x= \frac {a}{c}$ , $(x- \frac {3+\sqrt 5}{2})(x- \frac {3- \sqrt 5}{2})$ , $(\frac {3- \sqrt 5}{2}) <x <(\frac {3+\sqrt 5}{2})$ . Now obviously the upper bound cannot be improved,and holds for a degenerate triangle, and so we take $C= \frac {3+\sqrt 5}{2}$ , so that the answer to the question is $8$ .

The condition in the given question states that a/b = b/c. This means that

a/c = (b^2)/(c^2). ----- (1)

By triangle inequality, we have a < b+c. Dividing both sides by c, we get

a/c = 1+ (b/c). ----- (2)

Since there is no further specifications on the dimensions of the triangle. We will rely on the aforementioned equations to solve the problem.

(1) and (2) gives us

(b^2)/(c^2) < (b/c) + 1

(b^2)/(c^2) - (b/c) - 1 < 0

Solving for b/c and squaring to find a/c, we have

a/c < (3 + 5^(1/2))/2 C = (3 + 5^(1/2))/2 2C = (3 + 5^(1/2))

Therefore, a+b = 3+5 = 8.

In this problem, we have to find the maximum value of $\frac{a}{c}$ , subject to the condition $\frac{a}{b} = \frac{b}{c}$ , where a, b and c are the sides of a triangle. Clearly, b has value intermediate to those of a and c, and to maximize $\frac{a}{c}$ , we have to set c > a.

Let $b=kc$ , $a=k^{2}c$ , where $k>1$ .

We know that sum of two sides is greater than the third, and more specifically, the greatest side is smaller than the sum of the other two sides. Hence, $k^{2}c \leq kc + c$

i.e., $k^{2} - k -1 \leq 0$

i.e. k being +ve, $k \leq \frac{\sqrt{5}+1}{2}$

Hence $C = k_{max}^{2} = \frac{3+\sqrt{5}}{2}$

WLOG, let $a = 1$ . Then, $c = b^{2}$ . By triangle inequality, $b^{2} - b - 1 < 0$ and $b^{2} + b - 1 > 0$ . Solving these inequalities gives $\frac{\sqrt{5} - 1}{2} < b < \frac{\sqrt{5} + 1}{2}$ . We want to find the maximum possible value of $\frac{1}{b^{2}}$ . This is equal to the value of $\frac{4}{(\sqrt{5} - 1)^{2}} = \frac{3 + \sqrt{5}}{2}$ . The answer is $3 + 5 = 8$ .

Without loss of generality, we may take $c = 1$ , since all quantities of interest ( $a/b$ , $b/c$ , $a/c$ ) are expressed as ratios of the side lengths. Note that the equilateral triangle $a = b = c = 1$ is in $S$ : $a/b = b/c = 1$ , hence $C \ge M_{ABC} = a/c = 1$ . So we need only consider the case $a > b > c$ , which implies $a = b^2$ . The only triangle inequality not already trivially satisfied is $b+c > a$ , or $b^2 - b - 1 < 0$ , and solving the quadratic gives the upper bound $1 < b < \frac{1 + \sqrt{5}}{2}.$ Thus the least upper bound of $a/c = a = b^2$ is $C = \left( \frac{1 + \sqrt{5}}{2} \right)^2 = \frac{3+\sqrt{5}}{2},$ and $2C = 3 + \sqrt{5}$ , so the answer is $8$ .

Let $\frac{a}{b}=\frac{b}{c}=k$ , thus $ck^2=bk=a$ . Note $a, b, c, k>0$ .

Since $\frac{a}{c}$ is maximized, we need a to be the longest side and c to be the shortest side, instead of c being the longest side and a being the shortest side. Note that a, c cannot be the median side (unless k=1) as:

$k>1 \Rightarrow a>b>c$ or $k<1 \Rightarrow a<b<c$ .

Since a is the longest side now, we can use the triangle inequality.

$a<b+c \Rightarrow ck^2<ck+c \Rightarrow k^2-k<1$

Since the function $f(k)=k^2-k$ is increasing, thus we need $k<$ positive root of $k^2-k-1$ . By the quadratic formula,

$x^2-x-1=0 \Rightarrow x=\frac{1 \pm \sqrt{5}}{2}$

The positive root is thus $\frac{1+\sqrt{5}}{2}$ , thus $k<\frac{1+\sqrt{5}}{2}$

Since both sides are positive (First line, k>0), we can square both sides of the inequality.

$\frac{a}{c}=k^2<\frac{3+\sqrt{5}}{2}$

$2C=2 max{k^2}=3+\sqrt{5}$

Thus $a+b=8$

given: a/b=b/c=r(let)

so sides of triangle are b/r, b ,br

b/r + b > br

r^2-r-1>0 which gives {-1 + sqrt{5} } / { 2}<r< {-1 + sqrt{5} } / { 2}

max. value of M_{ABC}=r^2 < {3 + sqrt{5} } / { 2}

so 2*C={3 + sqrt{5} } / { 2}

C= 3 + sqrt{5}

     so required ans=3+5=8.

Let us suppose a/b=b/c=k. Therefore, we can get a=bk=ck^2 (k>1). Since the sum of the lengths of any two sides of a triangle must be greater than the third side, a<b+c which means ck^2<ck+c Therefore, by quadratic formula, we will get 1<k<(sqrt5+1)/2 Notice that C≥M=a/c=k^2 Since the largest value of k^2<[(sqrt5+10/2]^2=(3+sqrt5)/2, easily we can conclude that the value of 2C will be 3+sqrt5. Hence, the value of a+b=8

Say we have a triangle $ABC$ of $S$ , with side lengths $a, b, c$ . The number $M_{ABC}$ remains unchanged if we divide all the side lengths by $c$ , and it is still a triangle of $S$ , so we may assume without loss of generality that $c = 1$ . In that case, by definition of the set $S$ , we have $\frac{a}{b} = b$ , which gives $a = b^2$ . We also have $M_{ABC} = \frac{a}{c} = b^2$ .

We have now reduced the problem to looking at triangles with side lengths $1, b, b^2$ , and the problem is asking us for the supremum $C$ of the possible values of $b^2$ , with the only limiting factor being that $1, b, b^2$ are the sides of a triangle. If we define $s = \sqrt{C}$ , then $s$ is the supremum of values of $b$ , with the limitation that $b$ statisfies the triangle inequality: $b^2 < 1 + b$ We find that supremum by solving the equation $s^2 = 1 + s\\ \Downarrow\\ s = \frac{1\pm \sqrt{\left((-1)^2 - 4\cdot 1 \cdot (-1)\right)}}{2} \%this square root won't render properly on my computer, the parentheses partly solved the problem, but it's not pretty. = \frac{1 \pm \sqrt{5}}{2}$ Since we're only interested in the positive solution, we're left with $s = \frac{1+\sqrt{5}}{2}$ . Now we can calculate the number the problem asked us to find: $2C = 2s^2 = 2\cdot\frac{\left(1+\sqrt{5}\right)^2}{4} = \frac{1+2\cdot 1 \cdot\sqrt{5} + 5}{2} \\ = \frac{6 + 2\sqrt{5}}{2} = 3 + \sqrt{5}$ Now we have the number we were asked to find, on the form we were asked to write it, so the final answer is $3+5 = 8$ .

According to the Triangle inequality:

$b + c > a$

$\Leftrightarrow \frac{b}{c} + \frac{c}{c} > \frac{a}{c}$

$\Leftrightarrow \frac{b}{c} + 1 > \frac{a}{b}\cdot\frac{b}{c} (b,c>0)$

$\Leftrightarrow (\frac{b}{c})^2 - \frac{b}{c} - 1 < 0$

(Note that a, b, c are positive and so is $\frac{b}{c}$ )

$\Leftrightarrow \frac{b}{c} < \frac{1+\sqrt{5}}{2}$

$\Leftrightarrow M_{ABC} = \frac{a}{c} = (\frac{b}{c})^2 < \frac{6+2\sqrt{5}}{4}$

Thus, $2C = 3+\sqrt{5}$ and the value of $a+b$ is $8$

Note: other inequalities are then satisfied.

Let $\frac{a}{b}=\frac{b}{c}=r$ , then $M_{ABC}=\frac{a}{c}=\frac{a}{b}.\frac{b}{c}=r^2$

Without loss of generality, assume $\frac{a}{b}=\frac{b}{c}=r\geq 1$ , or $a\geq b\geq c$

By triangle's inequality, $a\leq b+c$ $\rightarrow \frac{a}{c}\leq \frac{b}{c} +1$ $\rightarrow r^2\leq r+1$ $\rightarrow (r-\frac{1}{2})^2\leq \frac{5}{4}$ $\rightarrow r \leq \frac{\sqrt{5}+1}{2}$ $\rightarrow r^2 \leq \frac{3+\sqrt{5}}{2}$

Therefore, $M_{ABC}=r^2\leq \frac{3+\sqrt{5}}{2} = C$ and $2C=3+\sqrt{5}$ . The desired answer is $3+5=8$ .