Math Circle Lecture | Ages 13–15 | Duration: 90 minutes | Topics: Geometry, Area, Similar Triangles, Trigonometry, Calculus (optional)
Triangle $ABC$ has base $AB = 70$ cm, left side $AC = 65$ cm, and right side $BC = 75$ cm. Find the position and length of a line perpendicular to the base $AB$ that divides the triangle into two regions of equal area.
Before we dive in, let's pause: Does such a line always exist? Could there be more than one?
We need the height from $C$ down to base $AB$. Let $F$ be the foot of that altitude.
Using Heron's Formula:
Semi-perimeter:
$$s = \frac{70 + 65 + 75}{2} = 105$$Area:
$$\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{105 \times 35 \times 40 \times 30}$$| Step | Computation | Value |
|---|---|---|
| $105 \times 35$ | $3{,}675$ | |
| $40 \times 30$ | $1{,}200$ | |
| Product | $3{,}675 \times 1{,}200$ | $4{,}410{,}000$ |
| Square root | $\sqrt{4{,}410{,}000}$ | $\mathbf{2{,}100}$ |
Now the height:
$$h = \frac{2 \times \text{Area}}{AB} = \frac{2 \times 2{,}100}{70} = 60 \text{ cm}$$Where does the altitude meet $AB$? Using the Pythagorean theorem in sub-triangle $AFC$:
$$AF^2 + h^2 = AC^2 \implies AF^2 = 65^2 - 60^2 = 4{,}225 - 3{,}600 = 625$$ $$AF = 25 \text{ cm}$$Check: $FB = 70 - 25 = 45$, and $45^2 + 60^2 = 2{,}025 + 3{,}600 = 5{,}625 = 75^2$ ✓
| Point | Coordinates |
|---|---|
| $A$ | $(0,\; 0)$ |
| $B$ | $(70,\; 0)$ |
| $C$ | $(25,\; 60)$ |
| $F$ (altitude foot) | $(25,\; 0)$ |
Our vertical line sits at some position $x = d$ on the base. But which slanted side does it meet — $AC$ or $BC$?
The altitude at $F$ splits the triangle into two parts:
| Sub-triangle | Base | Height | Area |
|---|---|---|---|
| Left ($AFC$) | $AF = 25$ | $60$ | $750$ |
| Right ($FBC$) | $FB = 45$ | $60$ | $1{,}350$ |
| Total | $2{,}100$ ✓ |
Half the total area is $\mathbf{1{,}050}$.
If the line were at $d \leq 25$, the left region would be a right triangle with legs $d$ and $\tfrac{12d}{5}$:
$$\text{Area} = \frac{6d^2}{5}$$Setting this to $1{,}050$ gives $d = 5\sqrt{35} \approx 29.6$, which is greater than 25. Contradiction!
For $25 < d < 70$, the vertical line at $x = d$ creates a right triangle on the right with vertices $P(d,\,0)$, $B(70,\,0)$, and $Q$ on side $BC$.
Height at position $d$: Side $BC$ descends from $C(25, 60)$ to $B(70, 0)$. Its slope is:
$$\text{slope of } BC = \frac{60 - 0}{25 - 70} = -\frac{4}{3}$$At horizontal distance $(70 - d)$ from $B$, the height is:
$$PQ = \frac{4}{3}(70 - d)$$Right triangle $BPQ$ has:
Set equal to half the total area:
$$\frac{2}{3}(70 - d)^2 = 1{,}050$$ $$(70 - d)^2 = 1{,}575$$ $$70 - d = \sqrt{1{,}575} = \sqrt{225 \times 7} = 15\sqrt{7}$$Verification: $\frac{1}{2}(15\sqrt{7})(20\sqrt{7}) = \frac{1}{2} \times 300 \times 7 = 1{,}050$ ✓
This is the most beautiful elementary approach.
Consider the right triangle $BFC$ (from the altitude):
Now consider our right triangle $BPQ$ (the cut piece):
By AA similarity: $\triangle BPQ \sim \triangle BFC$
The scale factor is:
$$k = \frac{BP}{BF} = \frac{70 - d}{45}$$Areas scale as the square of the linear ratio:
$$\frac{\text{Area}(\triangle BPQ)}{\text{Area}(\triangle BFC)} = k^2$$ $$\frac{1{,}050}{1{,}350} = k^2 \implies k^2 = \frac{7}{9} \implies k = \frac{\sqrt{7}}{3}$$Using the Law of Cosines:
$$\cos B = \frac{AB^2 + BC^2 - AC^2}{2 \cdot AB \cdot BC} = \frac{4{,}900 + 5{,}625 - 4{,}225}{10{,}500} = \frac{6{,}300}{10{,}500} = \frac{3}{5}$$So $\cos B = \frac{3}{5}$ and $\sin B = \frac{4}{5}$. A 3-4-5 angle!
In the right triangle $BPQ$, the hypotenuse $BQ$ lies along side $BC$:
$$BP = BQ\cos B = \tfrac{3}{5}\,BQ \qquad\qquad PQ = BQ\sin B = \tfrac{4}{5}\,BQ$$We need $\sin(2B)$:
$$\sin(2B) = 2\sin B\cos B = 2 \times \frac{4}{5} \times \frac{3}{5} = \frac{24}{25}$$Setting the area to $1{,}050$:
$$\frac{1}{4} \cdot BQ^2 \cdot \frac{24}{25} = 1{,}050 \implies \frac{6}{25}\,BQ^2 = 1{,}050 \implies BQ^2 = 4{,}375$$ $$BQ = \sqrt{4{,}375} = \sqrt{625 \times 7} = 25\sqrt{7}$$For students who know integration, we can compute the area to the left of $x = d$ directly.
At position $x$, the triangle's height above the base is:
$$y(x) = \begin{cases} \dfrac{12x}{5} & 0 \leq x \leq 25 \quad \text{(along side } AC\text{)} \\[8pt] \dfrac{4(70 - x)}{3} & 25 \leq x \leq 70 \quad \text{(along side } BC\text{)} \end{cases}$$First integral:
$$\int_0^{25} \frac{12x}{5}\,dx = \frac{6}{5}\,x^2\Big|_0^{25} = \frac{6}{5}(625) = 750$$(The area of sub-triangle $AFC$ — no surprise!)
Second integral:
$$\int_{25}^{d} \frac{4(70 - x)}{3}\,dx = \frac{4}{3}\left[70x - \frac{x^2}{2}\right]_{25}^{d} = \frac{4}{3}\!\left[\!\left(70d - \frac{d^2}{2}\right) - 1{,}437.5\right]$$Simplifying step by step:
$$70d - \frac{d^2}{2} - 1{,}437.5 = 225$$ $$d^2 - 140d + 3{,}325 = 0$$By the quadratic formula:
$$d = \frac{140 \pm \sqrt{19{,}600 - 13{,}300}}{2} = \frac{140 \pm \sqrt{6{,}300}}{2} = \frac{140 \pm 30\sqrt{7}}{2} = 70 \pm 15\sqrt{7}$$Since $d < 70$:
The other root $d = 70 + 15\sqrt{7} \approx 109.7$ lies outside the triangle — our quadratic "remembers" that the parabolic area function continues beyond the triangle's boundary.
| Quantity | Exact Value | Approximate |
|---|---|---|
| Position from $A$ | $70 - 15\sqrt{7}$ cm | $30.31$ cm |
| Position from $B$ | $15\sqrt{7}$ cm | $39.69$ cm |
| Length of line $PQ$ | $20\sqrt{7}$ cm | $52.92$ cm |
| Distance $BQ$ along $BC$ | $25\sqrt{7}$ cm | $66.14$ cm |
Each approach illuminates a different facet of the problem:
| Method | Key Idea | Prerequisites |
|---|---|---|
| 1 Direct computation | Height = slope × distance | Rise/run, Pythagorean theorem |
| 2 Similar triangles | Areas scale as $k^2$ | AA similarity, ratios |
| 3 Trigonometry | Use $\angle B$ via cosine rule | Law of Cosines, double-angle |
| 4 Integration | Area = integral of height | Definite integrals, quadratic formula |
All four arrive at the exact same answer — which is reassuring, and beautiful.
Instead of bisecting the area, cut off a fraction $r$ of the total area on the right. Show that:
$$70 - d = \sqrt{3{,}150\,r} \qquad\text{and}\qquad PQ = \frac{4}{3}\sqrt{3{,}150\,r}$$Start from $\frac{2}{3}(70 - d)^2 = 2{,}100\,r$ and solve for $(70-d)$.
What if the apex $C$ were at $x = 50$ instead of $x = 25$? Now the larger sub-triangle is on the left, and the bisecting line might cross side $AC$ instead. Work it out!
Is there a triangle where no perpendicular-to-base line can bisect the area? Think about what happens when the apex is directly above one endpoint of the base…
If $C$ is above $A$ (i.e., $AF = 0$), then the entire triangle lies on the right side of the altitude. A vertical line can still bisect it — the region to the left is always a right triangle. The equation $\frac{2}{3}(70 - d)^2 = 1050$ always has a solution in $(0, 70)$. So it's always possible! Every triangle can be bisected by a line perpendicular to any chosen base.
Find two vertical lines that divide our triangle into three regions of equal area ($700$ cm² each). Determine both positions.
The rightmost cut creates a right triangle of area $700$. The middle cut creates a region of area $700$ between the two cuts. Think carefully about whether both cuts are to the right of $F$, or one is to the left.
Math Circle Lecture — Geometry & Area Bisection