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- #1
- Feb 14, 2012
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Triangle ABC has \(\displaystyle AB=9\) and \(\displaystyle BC:AC=40:41\). What's the largest area that this triangle can have?
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Find the Largest Area of Triangle
- Thread starter anemone
- Start date
- Admin
- #2
My solution:
Let:
\(\displaystyle \overline{BC}=a\)
\(\displaystyle \overline{AC}=b\)
and we are told:
\(\displaystyle b=\frac{41}{40}a\)
Using Heron's formula, the area of the triangle as a function of $a$ is:
\(\displaystyle A(a)=\frac{9\sqrt{-81a^4+10499200a^2-207360000}}{6400}\)
The area is maximized where the radicand is maximized, hence:
\(\displaystyle \frac{d}{da}\left(-81a^4+10499200a^2-207360000 \right)=4a\left(5249600-81a^2 \right)=0\)
Discarding $a=0$, and taking the positive root, we find the critical value is:
\(\displaystyle a=\frac{40\sqrt{3281}}{9}\)
We know this is a maximum since the radicand is a quartic with a negative leading coefficient.
Thus, the maximum area is:
\(\displaystyle A_{\max}=A\left(\frac{40\sqrt{3281}}{9} \right)=820\)
\(\displaystyle \overline{BC}=a\)
\(\displaystyle \overline{AC}=b\)
and we are told:
\(\displaystyle b=\frac{41}{40}a\)
Using Heron's formula, the area of the triangle as a function of $a$ is:
\(\displaystyle A(a)=\frac{9\sqrt{-81a^4+10499200a^2-207360000}}{6400}\)
The area is maximized where the radicand is maximized, hence:
\(\displaystyle \frac{d}{da}\left(-81a^4+10499200a^2-207360000 \right)=4a\left(5249600-81a^2 \right)=0\)
Discarding $a=0$, and taking the positive root, we find the critical value is:
\(\displaystyle a=\frac{40\sqrt{3281}}{9}\)
We know this is a maximum since the radicand is a quartic with a negative leading coefficient.
Thus, the maximum area is:
\(\displaystyle A_{\max}=A\left(\frac{40\sqrt{3281}}{9} \right)=820\)
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- #3
- Feb 7, 2012
- 2,799
Here is a geometric way to get the same answer.
If the ratio $BC:AC$ is $40:41$ then $C$ must lie on an Apollonius circle with diameter $PQ$, where $P,Q$ lie on the line $AB$, with the ratios $BP:AP$ and $BQ:AQ$ both equal to $40:41$. Since $AB=9$, we must have $AP = 41/9$, $PB = 40/9$ and $BQ = 360.$ Thus the radius of the circle is $\frac12\bigl(360 + \frac{40}9\bigr) = 1640/9.$ The area of the triangle is obviously greatest when $C$ is at its maximum height above the line $AB$, in other words when it is directly above the centre of the circle. The height $OC$ is then the radius of the circle, and the area of the triangle is $\frac12AB*OC = \dfrac92\dfrac{1640}9 = 820.$
If the ratio $BC:AC$ is $40:41$ then $C$ must lie on an Apollonius circle with diameter $PQ$, where $P,Q$ lie on the line $AB$, with the ratios $BP:AP$ and $BQ:AQ$ both equal to $40:41$. Since $AB=9$, we must have $AP = 41/9$, $PB = 40/9$ and $BQ = 360.$ Thus the radius of the circle is $\frac12\bigl(360 + \frac{40}9\bigr) = 1640/9.$ The area of the triangle is obviously greatest when $C$ is at its maximum height above the line $AB$, in other words when it is directly above the centre of the circle. The height $OC$ is then the radius of the circle, and the area of the triangle is $\frac12AB*OC = \dfrac92\dfrac{1640}9 = 820.$
- Thread starter
- Admin
- #4
- Feb 14, 2012
- 3,932
Hi MarkFL and Opalg,
Thank you so much for participating in this problem.
Opalg, your solution is definitely an ingenious approach and I have talked to MarkFL and we don't think we would have thought of that geometric approach to solve this problem!
Thank you so much for participating in this problem.
Opalg, your solution is definitely an ingenious approach and I have talked to MarkFL and we don't think we would have thought of that geometric approach to solve this problem!
- Admin
- #5
Not so quick...where is your solution?Hi MarkFL and Opalg,
Thank you so much for participating in this problem.
Opalg, your solution is definitely an ingenious approach and I have talked to MarkFL and we don't think we would have thought of that geometric approach to solve this problem!
- Thread starter
- Admin
- #6
- Feb 14, 2012
- 3,932
My solution:
We first let the three sides of the triangle ABC be \(\displaystyle 9,\;40k,\;41k\).
By using the Heron's formula to find the area of the triangle ABC\(\displaystyle (A_{\text{triangle ABC}})\), we see that we have
\(\displaystyle A_{\text{triangle ABC}}=\sqrt{\left( \frac{81k+9}{2}\right)\left(\frac{81k+9}{2}-9\right)\left(\frac{81k+9}{2}-40k\right)\left(\frac{81k+9}{2}-41k\right)}\)
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{3}{4} \sqrt{-81\left(k^4-\frac{(81^2+1)k^2}{81}+1\right)}\)
And if we let \(\displaystyle k^2=x\), the equation above becomes
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{3}{4} \sqrt{-81\left(x^2-\frac{(81^2+1)x}{81}+1\right)}\)
By completing the square of the radicand, we obtain
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{3}{4} \sqrt{-81\left(x-\frac{3281}{81}\right)^2+\frac{3281^2}{81}-81}\)
Hence,
\(\displaystyle A_{\text{maximum}}=\frac{3}{4} \sqrt{\frac{3281^2}{81}-81}=820\)
Alternatively, we could approach this problem by applying AM-GM inequality if we write the radicand a bit differently.
\(\displaystyle A_{\text{triangle ABC}}=\sqrt{\left( \frac{81k+9}{2}\right)\left(\frac{81k+9}{2}-9\right)\left(\frac{81k+9}{2}-40k\right)\left(\frac{81k+9}{2}-41k\right)}\)
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{1}{4} \sqrt{ 9(9k+1)(9k-1)(9+k)(9-k)}\)
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{1}{4} \sqrt{ (9k+1)(9k-1)(81+9k)(81-9k)}\)
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{1}{4} \sqrt{ (81k^2-1)(81^2-81k^2)}\)
By applying AM-GM inequality to these two terms \(\displaystyle 81k^2-1\) and \(\displaystyle 81^2-81k^2\) gives
\(\displaystyle \frac{81k^2-1+81^2-81k^2}{2}\ge \sqrt{(81k^2-1)(81^2-81k^2)}\)
\(\displaystyle \sqrt{(81k^2-1)(81^2-81k^2)} \le \frac{81^2-1}{2}\)
Now, we can conclude that
\(\displaystyle A_{\text{maximum}}=\frac{1}{4}\left( \frac{81^2-1}{2}\right)=820\)
We first let the three sides of the triangle ABC be \(\displaystyle 9,\;40k,\;41k\).
By using the Heron's formula to find the area of the triangle ABC\(\displaystyle (A_{\text{triangle ABC}})\), we see that we have
\(\displaystyle A_{\text{triangle ABC}}=\sqrt{\left( \frac{81k+9}{2}\right)\left(\frac{81k+9}{2}-9\right)\left(\frac{81k+9}{2}-40k\right)\left(\frac{81k+9}{2}-41k\right)}\)
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{3}{4} \sqrt{-81\left(k^4-\frac{(81^2+1)k^2}{81}+1\right)}\)
And if we let \(\displaystyle k^2=x\), the equation above becomes
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{3}{4} \sqrt{-81\left(x^2-\frac{(81^2+1)x}{81}+1\right)}\)
By completing the square of the radicand, we obtain
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\frac{3}{4} \sqrt{-81\left(x-\frac{3281}{81}\right)^2+\frac{3281^2}{81}-81}\)
Hence,
\(\displaystyle A_{\text{maximum}}=\frac{3}{4} \sqrt{\frac{3281^2}{81}-81}=820\)
Alternatively, we could approach this problem by applying AM-GM inequality if we write the radicand a bit differently.
\(\displaystyle A_{\text{triangle ABC}}=\sqrt{\left( \frac{81k+9}{2}\right)\left(\frac{81k+9}{2}-9\right)\left(\frac{81k+9}{2}-40k\right)\left(\frac{81k+9}{2}-41k\right)}\)
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{1}{4} \sqrt{ 9(9k+1)(9k-1)(9+k)(9-k)}\)
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{1}{4} \sqrt{ (9k+1)(9k-1)(81+9k)(81-9k)}\)
\(\displaystyle \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \frac{1}{4} \sqrt{ (81k^2-1)(81^2-81k^2)}\)
By applying AM-GM inequality to these two terms \(\displaystyle 81k^2-1\) and \(\displaystyle 81^2-81k^2\) gives
\(\displaystyle \frac{81k^2-1+81^2-81k^2}{2}\ge \sqrt{(81k^2-1)(81^2-81k^2)}\)
\(\displaystyle \sqrt{(81k^2-1)(81^2-81k^2)} \le \frac{81^2-1}{2}\)
Now, we can conclude that
\(\displaystyle A_{\text{maximum}}=\frac{1}{4}\left( \frac{81^2-1}{2}\right)=820\)
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