A 2023 Math Kangaroo Problem Solved Two Ways

There's a particular category of geometry problem that looks, at first glance, like it's missing information. You stare at it. You count what you've been given. Two measurements, a circle, some squares. Surely that's not enough? And then the machinery clicks into place, and you realize the diagram was never incomplete—you just hadn't seen what it was constraining yet.

That's the experience Presh Talwalkar engineers in a recent MindYourDecisions video, working through a problem from the 2023 Math Kangaroo competition—a international mathematics contest for students that runs at multiple grade levels. The problem is clean, the solution is elegant, and Talwalkar does something worth paying attention to: he solves it twice, using different tools, to arrive at the same answer. That choice isn't just pedagogical generosity. It's a quiet argument about how mathematical understanding actually works.

The Setup

Here's the problem as Talwalkar presents it: a large square is divided into four equal smaller squares. A circle is drawn tangent to the midpoint of the large square's right side. The circle intersects three sides of the smaller squares. In the upper-left small square specifically, two measurements are given—the circle intersects the vertical side of that small square at a point 6 cm from the top of the large square, and it intersects the horizontal side at a point 8 cm from the left side of the large square. Find the side length of the large square.

No radius given. No center coordinates. Just two intersection distances and a tangency condition. It's the kind of constraint that competition problems are built on: seemingly sparse, actually overdetermined enough to pin a unique solution.

Talwalkar sets the side length of each small square as S, which makes the large square's side length 2S. The goal is to find 2S.

Method One: The Chord-Chord Intersection Theorem

The first approach leans on a theorem that's elegant precisely because it feels slightly surprising the first time you encounter it. The chord-chord intersection theorem states: when two chords of a circle intersect inside the circle, the product of the two segments of one chord equals the product of the two segments of the other.

In this problem, the circle's horizontal chord—the one passing through the tangency point on the right side—and the vertical chord through the upper-left small square intersect inside the circle. Talwalkar carefully labels the segments.

Because the circle is tangent to the midpoint of the large square's right side, the horizontal chord is perpendicular to that side. That perpendicularity means the horizontal chord is actually a diameter—a fact that becomes load-bearing in method two, but here just helps establish segment lengths. Working through the geometry, the horizontal chord's two segments measure (S − 8) and S. The vertical chord, using the symmetry created by that perpendicular bisector relationship, splits into two equal segments of (S − 6) each.

Plug those into the chord-chord theorem:

(S − 8) × S = (S − 6) × (S − 6)

Expand both sides: S² − 8S = S² − 12S + 36

The S² terms cancel. What remains is a clean linear equation: 4S = 36, so S = 9.

The large square's side is 2S = 18 cm.

As Talwalkar puts it: "I think using the chord-chord intersection theorem is the most direct way to solve this problem." Hard to argue. The algebra is almost embarrassingly tidy once you've set up the segments correctly.

Method Two: Pythagorean Theorem and a Constructed Triangle

Here's where the video earns its second half. Talwalkar acknowledges that not everyone will remember the chord-chord theorem on demand—particularly under competition conditions—and offers a route that relies only on the Pythagorean theorem and the geometric consequences of the tangency condition.

Since the horizontal chord is a diameter (established by the perpendicularity argument), its length is (S − 8) + S = 2S − 8, making the radius R = S − 4.

Now Talwalkar constructs a right triangle. The right angle sits at the intersection of the two chords. One leg runs from that intersection point to the end of the diameter along the horizontal, a distance of 4 (since the full half-diameter is S − 4, and the segment from the intersection to the tangency point on the right side is S, making the remaining piece 4). The other leg runs along the vertical chord, length (S − 6). The hypotenuse connects the intersection point to a point on the circle, so it's a radius: (S − 4).

Pythagorean theorem:

4² + (S − 6)² = (S − 4)²

16 + S² − 12S + 36 = S² − 8S + 16

The 16s cancel. The S² terms cancel. You're left with −12S + 36 = −8S, which gives 4S = 36, and again S = 9.

Same answer. Different path through the same geometric landscape.

"This method is going to capitalize on the fact that these two chords are perpendicular to each other," Talwalkar explains—and that perpendicularity, which felt like a minor detail in the setup, turns out to be what makes the whole second approach possible.

What Two Solutions Actually Tell You

This is where I want to sit for a moment, because the pedagogical choice here is more interesting than it might appear.

In mathematics education, there's a persistent tension between efficiency and understanding. Teaching the most powerful theorem for a given class of problems is efficient—students can solve more problems faster. But it can also produce a kind of brittle fluency: students who can apply chord-chord correctly whenever they recognize the setup, but who don't have a feel for why it works or what they'd do if they couldn't pattern-match to it.

Talwalkar's two-method structure implicitly addresses this. The Pythagorean approach is slower and requires more geometric construction, but it forces you to actually see the circle's structure—where the center is, what the radius constrains, how the tangency condition propagates through the diagram. The chord-chord method is fast, but it works somewhat opaquely unless you already understand what the theorem is encoding (which is, at root, a consequence of similar triangles formed by intersecting chords).

Seeing both routes converge on S = 9 isn't just reassuring—it's informative. It tells you something about the problem's structure that neither method reveals alone.

Math Kangaroo, for its part, is designed with exactly this kind of thinking in mind. The competition spans grade levels from 1 to 12, and its problems are chosen to reward insight over rote procedure. A problem like this one—solvable by multiple theorems, but only if you've correctly parsed what the diagram is actually constraining—fits that philosophy neatly.

The Geometry Behind the Numbers

It's worth pausing on the specific numbers the problem uses: 6 and 8. Those aren't arbitrary. Notice that in the final solution, S = 9, which means the vertical chord segments are each S − 6 = 3, and the horizontal chord segments are S − 8 = 1 and S = 9. Check the chord-chord product: 1 × 9 = 9, and 3 × 3 = 9. Clean.

In the Pythagorean method, the triangle has legs of 4 and 3, and a hypotenuse of 5. That's a 3-4-5 right triangle—the most recognizable Pythagorean triple in existence. Whether this was intentional on the part of the problem designers is unknowable from the outside, but it's hard to believe it's accidental. Competition problem design tends to be precise about the arithmetic experience it creates. A 3-4-5 triangle lurking inside a problem about squares and circles is exactly the kind of hidden structure that makes a problem feel right when it resolves.

That said, the 6 and 8 in the original setup also hint at another famous triple: 6, 8, 10—a scaled 3-4-5. Whether that's a deliberate planted signal or a coincidence of the problem's constraints is the sort of question that mathematicians find genuinely fun to argue about, and that I find genuinely fun to leave open.

"What a fun problem," Talwalkar says at the end, and it reads less like a closing pleasantry than like a considered verdict. The 2023 Math Kangaroo competition apparently agreed—it chose this problem for a reason.

The real question it leaves behind isn't about the answer, which is 18 cm and not in serious dispute. It's about which theorem you reach for first, and what that choice reveals about how you've learned to see a diagram.

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By Amelia Nwofor, Science Desk Editor