Area Inside an Obtuse Triangle

From the problem, we can see that $AEDF$ is a parallelogram. Hence, $[DEF] = [AEDF] / 2$ . Next, note that $CED \sim CAB$ and $BDF \sim BCA$ . Furthermore, since $BD+DC=BC$ , their proportionality constants add up to 1. Let them be $x$ and $1-x$ respectively. Then, $[AEDF] = [ABC] - [BDF] - [CDE] = [ABC] (1-x^2-(1-x)^2)$ . By Power Mean Inequality, $x^2+(1-x)^2 \geq 1/2$ . Thus, $[AEDF] \leq [ABC] / 2$ , and $[DEF] \leq [ABC] / 4$ . We may easily compute $[ABC] = 360$ , so $[DEF] \leq 90$ . Equality is attained when $1-x = x, x=1/2$ , and $D$ is selected to be the midpoint of $BC$ . $DEF$ is known as the medial triangle of $ABC$ .

We note that the area of a triangle is $\frac{ab\,\sin C}{2}$ , so $[ABC]=\frac{(AB)(AC)\, \sin\angle BAC}{2}=360$ and

$[DEF]=\frac{(DE)(DF)\,\sin \angle FDE}{2}$ . Thus, it suffices to maximize

$360\cdot\frac{[DEF]}{[ABC]} =360\cdot\frac{DE}{AB}\cdot\frac{DF}{AC}\cdot\frac{\sin\angle FDE}{\sin\angle BAC}\quad(1)$ .

Since $DE\parallel AB$ and $DF\parallel AC$ , we know that $AEDF$ is a parallelogram, and $\angle FDE=\angle BAC\implies\sin\angle FDE=\sin\angle BAC$ . Furthermore, $\triangle BFD\sim\triangle BAC$ and $\triangle CDE\sim\triangle CBA$ . Let us set $BD=x$ , which in turn, implies that $CD=30-x$ . Using these similar triangles, we can see that $\frac{DF}{AC}=\frac{x}{30}$ and $\frac{DE}{AB}=\frac{30-x}{30}$ .

When we substitute these ratios into $(1)$ , we see that we need to maximize $360\cdot\frac{30-x}{30}\cdot\frac{x}{30}\cdot1=\frac{-2x^2+60x}{5}$ . The maximum value of this quadratic expression occurs at $x=-\frac{60}{2(-2)}=15$ . We note that at the maximum, $D$ is the midpoint of ${BC}$ , $DE\parallel AB$ , and $DF\parallel AC$ , so $\triangle DEF$ is the medial triangle of $\triangle ABC$ , and has $\frac{1}{4}$ the area, so $\max([DEF])=\frac{1}{4}\cdot 360=\boxed{90}$ .

First, draw the figure and determine the angles. Now assume $DC=x, BD=b-x$ , where $b=BC=30$ . From $\triangle DEC$ , we can write $ED=\frac{\sin{C}}{\sin{A}}x$ and from $\triangle FDB$ we can write $FD=\frac{\sin{B}}{\sin{A}}(b-x)$ . Hence $[DEF]=\frac{1}{2}DE\times DF \sin \angle FDE$ $=\frac{1}{2}\frac{\sin{C}\sin{B}}{\sin{A}}x(b-x)$

Clearly, $[DEF]$ is maximized at $x=\frac{b}{2}=15$ . And $[DEF]_{\max}$ $=\frac{b^2}{8}\frac{1}{\cot{C}+\cot{B}}$ (by writing $\sin{A}=\sin({B+C})$ and dividing the Dr. by the Nr.) $= \frac{30\times 24}{8}=90$

Let $a=\frac{BD}{BC}$ . Since $DEF$ is congruent to $AFE$ , we have $[ABC]=[FBD]+2[DEF]+[EDC]$ , which implies that $[DEF]=\left(\frac{1-a^2-(1-a)^2}{2}\right)[ABC]=a(1-a)[ABC]$

As $D$ varies on $BC$ , this allows $a$ to vary in $[0,1]$ while $[ABC]$ remains fixed. Thus the maximum occurs when $a(1-a)$ is maximized, which is at $a=\frac{1}{2}$ . Then $[DEF]=\frac{1}{4}[ABC]=90$

We let the ratio BD/BC be r. To maximize the area of DEF, we will find the optimal value of r. Note that triangle EDC is similar to triangle ABC. We then have that ED/DC = AB/BC, which yields that DE = (1-r)AB. We now drop a perpendicular from B to DE, and call the intersection G. Note that BGD will stay the same shape no matter where D is chosen. If we let BG/BD = k, we then have that BG = (kr)BC. The area of triangle DEF, from the base-height formula, is DE BG/2 = AB BC kr(1-r)/2. Since AB, BC, and k are fixed, we need to maximize r(1-r), which occurs at r = 1/2. Thus, D is the midpoint of BC. From the midpoint-parallel line theorem, E and F are the midpoints of AC and AB, respectively. Note that triangles AFE, DEF, FBD, and EDC are all congruent. Thus, the area of DEF is one-fourth that of ABC. The area of ABC, using the base-height formula, is 30 24/2 = 360. Dividing by four to get the area of DEF yields 90.

First, notice that $[\bigtriangleup DEF]$ = $\frac{[ AFDE]}{2}$ . So as we know that $[\bigtriangleup ABC]=\dfrac{24\times 30}{2}=360,$ it remains to find the minimum of $[\bigtriangleup FBD]+[\bigtriangleup EDC]$ .

First, since $DE\parallel AB, FD\parallel AC$ , we know that $\dfrac{CE}{CA}=\dfrac{CD}{CB}, \dfrac{BD}{BC}=\dfrac{BF}{BA}.$ So $[\bigtriangleup BFD]=(\dfrac {BD}{BC})^{2}\times [\bigtriangleup ABC]=360(\dfrac {BD}{30})^{2}.$ Similarly, $[\bigtriangleup CED]=(\dfrac {CD}{BC})^{2}\times [\bigtriangleup ABC]=360(\dfrac {30-BD}{30})^{2},$ since $CD=BC-BD=30-BD.$ So $[\bigtriangleup FBD]+[\bigtriangleup EDC]=360((\dfrac {BD}{30})^{2}+(\dfrac {30-BD}{30})^{2}).$

Now we investigate the minimum of $(\dfrac {BD}{30})^{2}+(\dfrac {30-BD}{30})^{2}$ . Obseve that by QM-AM inequality, since both $BD, 30-BD\ge 0,$ , we have: $\sqrt{\dfrac{{(BD)^{2}+(30-BD)^{2}}}{2}}\ge\dfrac {BD+30-BD}{2}=15,$ so $(BD^{2}+(30-BD)^{2}\ge 15^{2}\times 2=450$ , with equality holds if and only if $BD=15$ . So $(\dfrac {BD}{30})^{2}+(\dfrac {30-BD}{30})^{2}\ge \frac{450}{900}=0.5,$ with equality holds iff $BD=15$ .

Finally, $[\bigtriangleup FBD]+[\bigtriangleup EDC]=360((\frac {BD}{30})^{2}+(\frac {30-BD}{30})^{2})\ge 360\times 0.5=180,$ we have: $[AEDF]=[\bigtriangleup ABC]-[\bigtriangleup BFD]-[\bigtriangleup CED]\le 360-180=180,$ and so $[\bigtriangleup DEF]\le \frac{180}{2}=90.$ By the proof above, the equality case can be achieved when $BD=DC=15$ .

Hence the answer is $90$ .

Given the relationships in the question, we can deduce the following: AE = FD, AF = ED, common side EF, angle EFA = angle FED --> [DEF] = [AFE] FBD ~ EDC

Considering the trigonometric formula for triangles (0.5)(a)(b)(\sin \theta) when theta is the angle formed by adjacent sides Area of triangle is maximum when the ab\sin \theta is maximum.

Since base BC is divided into 2 lengths (BD, DC), and that FBD ~ EDC, [DEF] would be at maximum when BD = DC as this maximizes the lengths of individual triangles.

Incidentally, this condition results in the 4 triangles having equal areas (considering alternate angles, it can be deduced that angle AFE, FBD, EDC and FED are equal)

Area ABC = 0.5 * base * height = 360

Since maximum value of [DEF] occurs when the area of triangles AEF, FBD, FED and EDC are equal , maximum value = 360/4 = 90

Suppose we let point G be on DE such that FG is parallel to BC. Let x be the length of BD. Since FB || GD and FG || BD, FBDG is a parallelogram, so FG = x.

Note that

[DEF] = [DGF] + [EFG]

[DEF] = (1/2 * x * (distance from E to FG)) + (1/2 * x * (distance from D to FG))

[DEF] = 1/2 * x * (distance from E to BD)

Since ED || AB, triangle CDE is similar to triangle CBA. Thus:

CD / CB = (distance from E to BC) / (distance from A to BC)

(30 - x) / 30 = (distance from E to BC) / 24

distance from E to BC = 24 * (30 - x) / 30 = 4 * (30 - x) / 5

[DEF] = 1/2 * x * (4 * (30 - x) / 5) = (2/5) * x * (30 - x)

[DEF] = (2/5) * (30x - x^2) = (2/5) * (225 - (225 - 30x + x^2))

[DEF] = (2/5) * (225 - (15 - x)^2)

To minimize 225 - (15 - x)^2, we let 15 - x = 0:

[DEF] = (2/5) * 225 = 90.

First, let us note the area of $ABC$ remains constant, as the measure of $BC$ and the height corresponding to itself remains constant.

The area of $ABC$ is $\frac{30\times 24}{2}=360$. Now, since $DE$ is parallel to $AF$ and $DF$ is parallel to $AE$, we have $AEDF$ is a parallelogram, and thus, the area of $DEF$ is equal to the area of $AEF$.

Then again, because of the parallel segments we mentioned previously, we may note

$\frac{AF}{AB}=\frac{DC}{BC}=\frac{CE}{AC}=1-\frac{AE}{AC}.$

Let $\frac{1}{2}+x=\frac{AF}{AB}$ and $\frac{AE}{AC}=\frac{1}{2}-x$ with $0\leq x\leq 1$. Let $s\left(XYZ\right)$ denote the area of triangle $XYZ$. We have

$\frac{s\left(AEF\right)}{s\left(ABC\right)}=\frac{AF\cdot AE\cdot \sin{\angle AEF}}{AB\cdot AC \cdot \sin{\angle BAC}}=\left(\frac{1}{2}+x\right)\left(\frac{1}{2}-x\right)=\frac{1}{4}-x^{2}\leq \frac{1}{4}$

and thus we have

$4s\left(AEF\right)\leq \left(ABC\right)=360$

so

$s\left(AEF\right)\leq 90$

and this can be achieved when $x=0$, when $D$ is midpoint of segment $BC$.

By the Side Splitter Theorem, $\triangle ABC ~ \triangle FBD ~ \triangle EDC$ .

A formula for the area of a triangle is also $[ABC] = \frac{1}{2} ab \sin C$ . Hence, if two triangles have an angle where the angle measures are the same and the sides adjacent to the first triangle are $a$ and $a$ and the sides adjacent to the second triangle are $ma$ and $na$ , the ratio of the area of the second triangle to the first triangle is $mn$

Let $CD/BC=a$ . Hence $DB/BC=1-a$ , $BF/AB=1-a$ , $AF/AB=a$ , $CE/AC=a$ , and $AE/AC=1-a$ .

Hence, $[DEF] = [ABC] - [BFD] - [DEC] - [FAE] = [ABC] (1 - (1-a)^2 - a^2 - (1-a)a) = [ABC] (a-a^2)$ .

$a-a^2 = -(a- \frac{1}{2})^2 - \frac{1}{4}$ so the maximum value of $a-a^2$ is $\frac{1}{4}$ .

Hence, $[DEF] = [ABC] (a-a^2) <= \frac{1}{4} [ABC] = \frac{1}{4} \frac{1}{2} (24)(30)=90$ .

Since DE || AB and FD || AC, ΔEDC ~ ΔFBD ~ ΔABC with base to height ratio 5:4. Also, AFDE is a parallelogram, thus, 2[DEF] = [AFDE].

In ΔEDC, let base CD = x with height $\frac {4}{5}$ x and area $\frac {2}{5}x^2$ .

In ΔFBD, base DB = (30-x) with height $\frac {4}{5}$ (30-x) and area $\frac {2}{5}(30-x)^2$ .

[DEF] = $\frac {1}{2}$ [AFDE] = $\frac {1}{2}$ ( [ABC] - [EDC] - [FBD] )

= $\frac {1}{2}$ ( $\frac {1}{2}$ (30)(24) - $\frac {2}{5}x^2$ - $\frac {2}{5}(30-x)^2$ )

= 180 - $\frac {1}{5}x^2$ - $\frac {1}{5}(900 - 60x + x^2)$

= $\frac {2}{5}(30x - x^2)$

= $\frac {2}{5}(225 - 225 + 30x - x^2)$

= $\frac {2}{5}(225 - (15 - x)^2)$

$\leq \frac {2}{5}(225)$

= 90

We get that the area is 360. Because of similar triangles, FD/AC + DE/AB = 1. [DEF] = 1/2 DE DF sin(A), so we wish to maximize DE*DF, which is equivalent to maximizing (FD/AC)(DE/AB) because AC and AB are constant. By AM-GM, (FD/AC)(DE/AB) is maximized when (FD/AC)=(DE/AB)=1/2, so DEF is the medial triangle and [DEF] = 1/4 * 360 = 90.

AE//DF, AF//ED => AEFD is a //ogram, => [EFD] = 0.5[AEFD].

FBD is similar to ABC because FD//AC. Similarly, EDC is similar to ABC.

let DC/BC be a.

[EFD] = 0.5[AEFD]

= 0.5([ABC] - [EDC] - [FBD]) =0.5([ABC] - a^2[ABC] - (1-a)^2[ABC]) =0.5 ABC = ABC = ABC =< 0.25[ABC] = 90 sq. units

We can calculate [DEF] by doing [ABC]-[AFE]-[BFD]-[EDC]. [ABC] = 360 because 30 24/2=360. Assume length of BD is x, length of CD is 30-x. Then [BFD] = (x^2/900) 360 using ratios, [EDC] = ((900-60x+x^2)/900) 360, and [AFE] = (x(30-x)/900) 360. So [ABC]-[AFE]-[BFD]-[EDC] = [DEF] = 360-360 (x^2-30x+900)/900. Then taking the derivative gives 720x-10800=0, x=15. So the maximum value of [DEF] is 360-360(15^2-30 15+900)/900 = 360-360*675/900 = 360-270 = 90.

If the drawing is correct then the triangles CED and CAB are proportional with the ratio $k=\frac {CD}{CB} = \frac {ED}{AB}$ and the triangles BDF and BCA are proportional with the ratio $1-k=\frac {BD}{CB} = \frac {FD}{AC}$ .

The area of DEF is $S=\frac {1}{2}\cdot DE\cdot DF\cdot sin(FDE)$ and because AFDE is a paralelogram the angles FDE and BAC have the same measure. Using the equations from the parragraph above, S (area of DEF) is $\frac {1}{2}\cdot kAB\cdot (1-k)AC\cdot sin(BAC)$ = $k(1-k) \cdot [ABC]=\frac {1}{2} \cdot k(1-k)BC \cdot H$ and H is the distance from A to the line BC.

Now, all we have to do is to maximize $k(1-k), 0 \leq k \leq 1$ . The maximum is reached for $k=0.5$ and the maximum is 0.25. Replacing k with 0.5, BC with 30, H with 24, the result is $S=\frac{0.25 \cdot 30 \cdot 24}{2}=90$

maximum area will be possible when D,E and f are midpoints of the sides BC,AC,AB respectively.therefore area =0.25\times 0.5\times 24\times 30=90

We use barycentric coordinates $A = (1,0,0), B = (0,1,0), C = (0,0,1), D = (0,p,1 - p), E = (1 - q,0,q), F = (r,1 - r,0)$ , where $p \in [0,1]$ and $q,r$ depend on $p$ , then $\overrightarrow{AB} = (-1,1,0)$ $\overrightarrow{AC} = (-1,0,1)$ $\overrightarrow{ED} = (q - 1,p,1 - p - q)$ $\overrightarrow{FD} = (-r,p + r - 1,1 - p)$ The fact that $DE \parallel AB$ and $FD \parallel AC$ gives us $\frac{q - 1}{p} = -1$ $\frac{-r}{1 - p} = -1$ Thus $r = 1 - p$ , $q = 1 - p$ , this gives us $[DEF] = [ABC] \cdot \begin{vmatrix} 0 & p & 1-p\\ p & 0 & 1-p\\ 1 - p & p & 0 \end{vmatrix} = [ABC] \cdot (1 - p)p = 360 \cdot (1 - p)p$ Hence we wish to maximize $360(1 - p)p$ , this is quadratic with roots $0,1$ , hence it approaches maximum at their arithmetic mean $p = 1/2$ , so the maximal area is $360 \cdot \frac{1}{2} \cdot \frac{1}{2} = \boxed{90} \text{.}$

If you draw a diagram of this problem, you find that $\delta DEF$ has $\frac {1}{4}$ of the area of $\delta ABC$ . Now, all you need to do is $24 \times 30 \times \frac {1}{2} \times \frac {1}{4}$ which gives a result of $90$ .