lesson
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If you snap a dry spaghetti noodle into three random pieces, can you always tape them into a triangle? Surprisingly, the answer is no — two short pieces might not even reach each other.
Before picking up your compass, the Triangle Inequality Theorem gives us a simple test: the sum of any two side lengths must be strictly greater than the third side (a+b>c).
📊Interactive visual comparing possible vs impossible triangle side lengths. Top card shows lengths 3 cm, 4 cm, 6 cm: two colored bars pivot from the ends of a base bar of length 6 cm and successfully meet at a vertex (3 + 4 = 7 > 6, 'Forms a Triangle!'). Bottom card shows lengths 2 cm, 3 cm, 6 cm: the two shorter bars pivot flat toward each other and leave an obvious gap (2 + 3 = 5 < 6, 'Cannot Close!'). Clean minimalist card layout with toggle buttons, #1e2945 text, #22c55e green for valid, #ef4444 red for impossible.
Once we know three segments can legally form a triangle, how do we lock down that exact third vertex using only an unmarked straightedge and a compass?
The SSS Construction Method
Around 300 BCE, the Greek mathematician Euclid proved in Proposition 22 of the Elements that three valid segments pinpoint a single, rigid triangle. We call this SSS (Side-Side-Side) construction.
First, draw a working line with your straightedge and use your compass to duplicate the longest segment, creating base endpoints A and B.
Next, open the compass to the second side length, place the point on A, and swing an arc above the baseline. Every point on this arc is that exact distance from A.
Finally, set the compass width to the third side length, place the needle on B, and swing a second arc until it crosses the first arc at point C.
📊Step-by-step animated diagram showing the SSS construction of triangle ABC. Frame 1: Three given segments of lengths a, b, c labeled at the top. Frame 2: Base segment AB of length c drawn on a baseline. Frame 3: Compass needle at A draws a blue circular arc of radius b. Frame 4: Compass needle at B draws an orange circular arc of radius a, intersecting the blue arc at point C. Frame 5: Straight segments AC and BC drawn, highlighting the completed triangle ABC. Step controls (Previous/Next) allow manual stepping. Clean vector look, 350px responsive width.
Why does this guarantee the exact right triangle? Because a circle is the set of all points at a fixed distance from its center, so the intersection point C is the only location that satisfies both side lengths at the same time.
Now that you see how the arcs lock the third vertex in place, what errors should you watch out for during an exam?