AIME 1985 · 第 6 题

Solution 1

Let the interior point be $P$, let the points on $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$ be $D$, $E$ and $F$, respectively. Let $x$ be the area of $\triangle APE$ and $y$ be the area of $\triangle CPD$. Note that $\triangle APF$ and $\triangle BPF$ share the same altitude from $P$, so the ratio of their areas is the same as the ratio of their bases. Similarly, $\triangle ACF$ and $\triangle BCF$ share the same altitude from $C$, so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are actually the same, and thus in the same ratio. As a result, we have: $\frac{40}{30} = \frac{124 + x}{65 + y}$ or equivalently $372 + 3x = 260 + 4y$ and so $4y = 3x+ 112$.

Applying identical reasoning to the triangles with bases $\overline{CD}$ and $\overline{BD}$, we get $\frac{y}{35} = \frac{x+y+84}{105}$ so that $3y = x + y + 84$ and $2y = x + 84$. Substituting from this equation into the previous one gives $x = 56$, from which we get $y = 70$ and so the area of $\triangle ABC$ is $56 + 40 + 30 + 35 + 70 + 84 = \Rightarrow \boxed{315}$.

Solution 2 (Mass Points)

We can solve this problem using mass points. First, denote the interior point as $P$, the foot of the cevian from $A$ as $A'$, the foot of the cevian from $B$ as $B'$, and the foot of the cevian from $C$ as $C'$. Notice how triangles $\Delta APC'$ and $\Delta BPC'$ share the same altitude from $P$. Because their areas are in the ratio $40:30=4:3$, then their bases $AC'$ and $BC'$ must also be in the ratio $4:3$. Next, notice how triangles $\Delta APB$ and $\Delta PBA'$ also share the same altitude from P. Their areas are in the ratio $40+30:35=70:35=2:1$, so the ratio $AP:PA'$ must also be $2:1$.

Now, we can apply mass points. First, assign vertex $B$ a weight of $1$. Then, we determine the weight of $A$ as $\frac{BC'}{AC'}\cdot W_{B}=\frac{3}{4}$, and by adding the weights of $A$ and $B$, we get $W_{C'}=\frac{7}{4}$. Applying the same to $\overline{AA'}$, we get that $W_{A'}=\frac{3}{2}$, $W_{P}=\frac{9}{4}$, $W_{C}=\frac{1}{2}$, and $W_{B'}=\frac{5}{4}$.

Finally, we can determine the area of $\Delta ABC$ by adding together the areas of all of the small triangles, and using side length ratios derived from mass points to determine the areas of the two unknown triangles.

$$[\Delta ABC]=40+30+35+84+[\Delta APB']+[\Delta CPA']$$ $$[\Delta ABC]=189+\frac{2}{3}\cdot [\Delta CPB']+2\cdot [\Delta BPA']$$ $$[\Delta ABC]=189+56+70$$ $$[\Delta ABC]=\boxed{315}$$ ~ Danny Wang (ThePeeps191)

~ Diagram By Daniel Wang (fp77)

Solution 3

Let the interior point be $P$ and let the points on $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$ be $D$, $E$ and $F$, respectively. Also, let $[APE]=x,[CPD]=y.$ Then notice that by Ceva's, $\frac{FB\cdot DC\cdot EA}{DB\cdot CE\cdot AF}=1.$ However, we can deduce $\frac{FB}{AF}=\frac{3}{4}$ from the fact that $[AFP]$ and $[BPF]$ share the same height. Similarly, $x=\frac{84CE}{EA}$ and $y=\frac{35DC}{BD}.$ Plug and chug and you get $xy=84\cdot 35\cdot \frac{3}{4}=2205.$ Then notice by the same height reasoning, $\frac{84}{x}=\frac{119+y}{x+70}.$ Clear the fractions and combine like terms to get $35x=5880-xy.$ We know $xy=2205$ so subtraction yields $35x=3675,$ or $x=105.$ Plugging this in to our previous ratio statement yields $\frac{84}{105}=\frac{4}{5}=\frac{119+y}{175},$ so $y=21.$ Basic addition gives us $105+84+21+35+30+40=\boxed{315}.$

-dchen

Solution 4

Let the interior point be $P$ and let the points on $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$ be $D$, $E$ and $F$, respectively. Then the cevians $AD,BF,CE$ are concurrent, so we can use Ceva's Theorem, letting $\frac{BD}{DC}=\frac{a}{b}$ and $\frac{CF}{FA}=\frac{c}{d}$. Notice that $\frac{AE}{EB}=\frac{[\Delta APE]}{[\Delta EPB]}=\frac43.$

$$\frac{4}{3}\cdot \frac{a}{b}\cdot \frac{c}{d}=1\implies \frac{d}{c}=\frac{a}{b}\cdot \frac{4}{3}.$$ We know that $[\Delta CPD]=35\cdot \frac ba$ and $[\Delta APF]=84\cdot \frac dc,$ so

$$[\Delta ABC] = 84+84\cdot \frac dc + 35\cdot \frac ba + 40+30+35 = \left(1+\frac ba\right)(40+30+35).$$ We will now solve for $\frac ba$:

$$84+84\cdot \frac ab\cdot \frac 43 + 35\cdot \frac ba = \frac ba\cdot 105.$$ $$12+16\cdot \frac ab + 5\cdot \frac ba = \frac ba\cdot 15$$ $$10\left(\frac ba\right)^2-12\cdot \frac ba-16=0$$ Factoring this gives $\left(\frac ba-2\right)\left(10\cdot \frac ba - 8\right)=0,$ so the area of $\triangle ABC$ is

$$\left(1+\frac ba\right)(40+30+35)=3\cdot 105=\boxed{315.}$$ ~ RubixMaster21

Video Solution by OmegaLearn

https://youtu.be/5jwD5UViZO8?t=300

~ pi_is_3.14

Note

This problem gives redundant information. The area of any one of the 4 triangles given can be derived from the other three triangles' areas

~ inaccessibles