The Missing Triangle Height

Suppose $a, b, c$ denote the three sides of the triangle, and let $t_a=63, t_b=147, t_c$ be the respective three heights. From the equality of the area of the triangle, we have $63\cdot a = 147\cdot b = t_c\cdot c$ . By rearranging we have $\frac{a}{b} = \frac{7}{3}$ and so denote $a=7n, b=3n$ for a real number $n$ .

By triangle inequality we have $b+c>a \Rightarrow c>4n$ . Thus, from the initial equation we have $t_c = \frac{147\cdot b}{c} = \frac{441\cdot n}{c} < 110.25$ . Since $t_c$ is an integer, $t_c\leq 110$ and this is achievable by setting $n=110, c=441$ . So the maximum value for the third height is $110$

Let the side length of the triangle corresponding to the altitude of length $63$ be $a$ and $b$ for $147$ . Note that $63a = 147b$ so $3a = 7b$ . Let $s$ be the last side length. Note that by the triangle inequality we have $s \ge \frac{4b}{3}$ . This implies the third altitude can have length at most $\frac{3}{4} \cdot 147 = 110.25$ . Thus if it is an integer we have its length is at most $110$ . It is clear that this is obtainable as the triangle inequality is not violated in this case, so we are done.

Let's called the missing height, $h$ like the picture below.

Alt text

Now we need "the maximum possible value for the third height". This should tell you that the last height, $h$ can't be just any number. In fact it is not bigger than a certain number. This should make you think about the triangle inequality . Let's keep that in mind and proceed.

Using the formula of the area of a triangle, we can write:

$\frac{1}{2}\times a \times 63= \frac{1}{2}\times b \times 147=\frac{1}{2}\times c \times h$ .

Or, $63a=147b=ch$ $\cdots (1)$ .

Notice from $(1)$ that in order to maximize $h$ , we'll have to minimize $c$ .

And from the relation $a=\frac{147}{63} b$ , it is clear that $a>b$ .

Now let's use the triangle inequality.

We have, $b+c \geq a$ .

After putting $a=\frac{147}{63} b$ and rearranging, we have $c \geq \frac{4}{3} b$ $\cdots (2)$ .

Now let's go back to $(1)$ and consider this:

$147b=ch$ .

Here, we can make a substitution from $(2)$ to get:

$147b\leq \frac{4}{3}\times bh$ .

After cancelling out the $b$ 's and rearranging we have:

$h\leq \frac{147\times 3}{4}$ or $110.25$ .

So the maximum value that $h$ could have is $110.25$ [and in that case we'll have a degenerate triangle $ABC$ , a triangle with zero area].

Finally notice that the problem states that $h$ has to be an integer. What is the largest integer that is less than or equal to $110.25$ ?

$\boxed{110}$ and that is our answer.

Let $p$ be the measure of the side perpendicular to the height measuring 63. Note that in any triangle with sides measuring $a_1, a_2$ and $a_3$ with corresponding heights measuring $h_1, h_2, h_3$ , the product $a_i \times h_i$ is constant for $i = 1, 2, 3$ since these are all equal to twice the area of the triangle. Hence, the measure of the side corresponding to height measuring 147 is $\frac {3p}{7}$ . Hence by triangle inequality, the third side is between $p - \frac {3p}{7} = \frac {4p}{7}$ and $p + \frac {3p}{7} = \frac {10p} {7}$ exclusive. Hence the measure of the third side is between $63p \times \frac {7}{3p} = 44.1$ and $63p \times \frac {7}{10p} = 110.25$ , exclusive. Hence maximum integer measure for the third side is 110.

Lemma : If two altitudes of a triangle have lengths $a_1$ and $a_2$ , then, if $a_1 < a_2$ , length of the remaining altitude $a_3$ is in the range $\dfrac {1}{\dfrac {1}{a_1} - \dfrac {1}{a_2}} < a_3 < \dfrac {1}{\dfrac {1}{a_1} - \dfrac {1}{a_2}}.$ Proof : For $k \in \{ 1, 2, 3 \}$ , consider the following. If we drop the altitude $a_k$ and we say that it makes a right angle with side $s_k$ , then, from the triangle inequality, we have: $s_1 + s_2 > s_3, \qquad s_2 + s_3 > s_1$ So: $s_1 - s_2 < s_3 < s_1 + s_2$ Also, since $a_k\cdot s_k=2A$ , where $A$ is the area of the triangle, we have: $s_k = \dfrac {2A}{a_k}.$ Thus, we have, from substitution: $\dfrac {2A}{a_1} - \dfrac {2A}{a_2} < \dfrac {2A}{a_3} < \dfrac {2A}{a_1} + \dfrac {2A}{a_2}.$ Dividing each side by $2A$ and reciprocating, we have our desired result. $\blacksquare$

Using the lemma for $a_1=63$ and $a_2=147$ , we take the upper bound and substitute to get $a_3 < 110.25.$ Thus, the largest possible integer value is $\fbox {110}$ .

Suppose that side triangle are $a, b, c$

and altitude of triangle are $t_a, t_b,t_c$

From area of triangle $L$ we have:

$a=\frac{2L}{t_a}$

$b=\frac{2L}{t_b}$

$c=\frac{2L}{t_c}$

From triangle inequality:

$a+b >c >a-b$

So, we get:

$\frac{1}{\frac{1}{t_a}+\frac{1}{t_b}}<c<\frac{1}{\frac{1}{t_a}-\frac{1}{t_b}}$

for integer $c$ maximum is

Max $c=\lfloor\frac{t_a \times t_b}{t_a-t_b} \rfloor=110$

Let $a=BC,b=CA,c=AB$ and $h_a,h_b,h_c$ denote the height of altitudes dropped from $A,B,C$ respectively. Let $h_a=63,h_b=147$ . We have to find the maximum value of $h_c$ . Now $ah_a=bh_b \Rightarrow 63a=147b \Rightarrow b=\frac{3a}{7}$ Applying cosine law we get $c^2=a^2+b^2-2.a.b.\cos C$ $\Rightarrow c=\frac{a}{7}\sqrt{58-42\cos C}$ . Now $ch_c=63a \Rightarrow h_c=\frac{441}{\sqrt{58-42\cos C}}$ . Since $\triangle ABC$ is non-degenrate (because it has positive altitude lenghs), we have $\cos C<1 \Rightarrow -42\cos C>-42$ $\Rightarrow \sqrt{58-42\cos C}>4$ $\Rightarrow \frac{441}{\sqrt{58-42\cos C}}<\frac{441}{4}$ $\Rightarrow h_c<110+\frac 14$ . So the maximun possible integer value of $h_c$ is $110$ .

Consider the triangle to have sides $a,b,c$ with them being the bases for heights 63, 147 and $h$ respectively. So as the triangle is the same the area is the same, therefore, $(1/2)63 a=(1/2)147 b=(1/2)hc$ , or

$63a=147b=hc$

From $63a=147b$ We note that $b:a=3:7$ So, let $b=3u, a=7u$

This implies that $441u=hc$ or $h=441\times\frac {u}{c}$

Now by triangle in equality we can get; $a+b>c, c+a>b, c+b>a$ or

$21u>c, c>-4u, c>4u$

Hence

$4u<c<21u$

$4<\frac{c}{u}<21$

Therefore

$\frac{1}{21}<\frac{u}{c}<\frac{1}{4}$

Hence,

$441\times\frac{1}{21}<441\times\frac{u}{c}<441\times\frac{1}{4}$

$21<441\times\frac{u}{c}<110.25$

$21<h<110.25$

Hence the maximum $h$ value is $110$

If two altitudes of a triangle a and b are known, we can calculate the range of values for the third altitude c using the following inequation:

1/(1/a+1/b)< c< 1/(1/a-1/b) where a is the smaller altitude.

1/(1/63+1/147) = 44.1

1/(1/63-1/147) = 110.25

So the largest integer in that range is 110.

If two altitudes of a triangle a and b are known, we can calculate the range of values for the third altitude c using the following equation:

1/(1/a+1/b)≤c≤1/(1/a-1/b) where a is the smaller altitude.

1/(1/63+1/147) = 44.1

1/(1/63-1/147) = 110.25

So the largest integer in that range is 110

Let the sides of the triangle be a, b, and c such that the heights 63 and 147 are facing the sides a and b. So we calculate the area of the triangle in 2 ways to get that 63a/2=147b/2 which reduces to 3a=7b. So we can let a=7k and b=3k. Then c>4k. Now let the last height be h. Then since 63a/2=ch/2 we reduce this to 441>4h. So since h is an integer the maximum value for h is 110.

Let the area of the triangle be $A$ and the third height be $h$ . Since we want to maximize this third height, clearly the side perpendicular to the height of length 63 is the longest. Thus, we have the inequality $\frac{A}{h}+\frac{A}{147}>\frac{A}{63}$ . Multiplying through gives $441A+3hA>7hA \Longrightarrow 441>4h$ . Since $h$ is integral, its maximum value is $110$ .

Let the side lengths of the triangle be $a$ , $b$ , and $c$ such that their corresponding heights are $63$ , $147$ , and $h$ , respectively. We want to maximize $h$ . We have that $63a = 147b = hc$ since all three are equal to twice the area of the triangle.

By the Triangle Inequality, we must have $b + c > a$ . We write $c$ and $a$ in terms of $b$ : $c = \frac{147b}{h}$ and $a = \frac{147b}{63}$ . Therefore, we must have $b + \frac{147b}{h} > \frac{147b}{63}$ . We can divide both sides by $b$ to obtain $1 + \frac{147}{h} > \frac{147}{63}$ . This simplifies to $\frac{147+h}{h} > \frac{147}{63}$ . Cross multiplying, we obtain $147h < 63h + 147 \cdot 63$ , or $84h < 147 \cdot 63$ . We can divide both sides by $21$ to get $4h < 147 \cdot 3 = 441$ . Because $h$ is an integer, its maximum possible value is $110$ .

First, let us denote 3 sides of the triangle as a,b,c.

$\frac {a * 147}{2} = \frac {b * 63}{2}$

$\frac {a}{b} = \frac {63}{147} = \frac{3}{7}$

$a = \frac{3}{7} b$

We need to show that $\frac {a+b+c}{2} \geq max(a, b, c)$

We need the smallest size of c, because height and c are inversely proportional ( Area = $\frac{height * c}{2}$ ). Since $b \geq a$ and we want to maximize our height, thus $b \geq c$ and hence $max(a,b,c) = b$ .

$\frac {c * height}{2} = \frac {b * 63}{2}$

$c = \frac {b * 63}{height}$

$\frac {\frac{3}{7} b+b+c}{2} \geq b$

$\frac{10}{7} b+ \frac {b * 63}{height} \geq 2b$

$\frac{7}{4} * 63 \geq height$

$110.25 \geq height$ , Hence the maximum integer height is 110.

Let the missing height be $h$ , and let the lengths of the sides containing the feet of the altitudes $63$ , $147$ , and $h$ be $a$ , $b$ , and $c$ , respectively. Then $Area = \frac{63a}{2} = \frac{147b}{2} = \frac{hc}{2}$ , so $\frac{b}{a}=\frac{63}{147}=\frac{3}{7}$ . Now, let $a=7x$ and $b=3x$ so that $c > a-b = 7x-3x=4x$ by the triangle inequality. Applying this inequality and plugging in $a=7x$ after solving for $h$ in the earlier equation gives $h = \frac{63a}{c} < \frac{63(7x)}{4x}=110.25$ , so $h=110$ is the maximum integer value of $h$ .

Let $h$ be the third height, then since the area is the same using any height you calculate from, we have $$63a=147b=hc$$ where $a, b, c$, are the side lengths to the respective sides of the altitudes. Now let us try to express the side lengths in terms of the altitudes. We try to make each expression equal to a multiple of all the altitudes. Any multiple of that also suffices so let $$a=7hk, b=3hk, c=441k$$ Now by the triangle inequality: $$3hk+441k>7hk\implies k<441/4$$ So the greatest possible integer value of $k$ is $\boxed{110}$. We quickly check that this satisfies the triangle inequality: $$3hk+7hk>441k \implies 1100>441$$ so we are sure of our answer.

$$ To find the possible sides of triangle, we simply add the two known sides together to get the highest value possible and subtract the two sides to get the lowest value possible. If $\,a$ , $\,b$ , and $\,c$ are the length sides of $\Delta \text{ABC}$ where $\,a$ and $\,b$ are known, then for $\,a>b$ the possible value of $\,c$ is $$ $a-b<c<a+b.$ $$ Let $\,t_a$ , $\,t_b$ , and $\,t_c$ be the length height of $\Delta \text{ABC}$ . Since the base or side of triangle is inversely proportional to its height, then the possible value of $\,t_c$ if $\,t_a$ and $\,t_b$ are known is $$ $\frac{1}{t_a}+\frac{1}{t_b}<\frac{1}{t_c}<\frac{1}{t_b}-\frac{1}{t_a}.$ $$ Thus, the maximum possible value for the third height are $$ $\frac{1}{t_c}<\frac{1}{t_b}-\frac{1}{t_a}=\frac{1}{147}-\frac{1}{63}=\frac{4}{441}$ $$ for $\,t_c \in \mathbb{Z}$ , it is equal to $\;\boxed{110}$ . $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$