{
  "format": "stapler-pack",
  "version": 1,
  "kind": "quiz",
  "meta": {
    "title": "Triangle Centers & the Euler Line",
    "author": "Staples Education",
    "description": "The four classical triangle centers and the Euler Line, plus midsegments, the geometric-mean altitude, and the Triangle Inequality. Converted from a worked tutoring whiteboard, where several of these were answered wrong in session — those wrong answers are the distractors here.",
    "created": "2026-08-06T04:14:56.198Z",
    "license": "Proprietary - Staples Education"
  },
  "payload": {
    "quiz": {
      "id": "triangle-centers",
      "title": "Triangle Centers & the Euler Line",
      "blurb": "The four classical centers — centroid, incenter, circumcenter, orthocenter — what each is built from, where each lives, and the line three of them share.",
      "meta": "19 questions · easy → hard",
      "categories": [
        "triangles-review",
        "properties-of-triangles"
      ],
      "questions": [
        {
          "id": "tct-001",
          "topic": "Triangle Centers",
          "difficulty": "easy",
          "stem": "Which triangle center is formed by the intersection of the three altitudes of a triangle?",
          "choices": [
            "Incenter",
            "Centroid",
            "Circumcenter",
            "Orthocenter"
          ],
          "correctIndex": 3,
          "explanations": [
            "The incenter is where the three ANGLE BISECTORS meet. Altitudes and angle bisectors coincide only in special triangles, so the two centers are generally different points.",
            "The centroid is where the three MEDIANS meet — the segments from each vertex to the midpoint of the opposite side. A median goes to a midpoint; an altitude goes in perpendicular. Those are different lines unless the triangle is isosceles about that vertex.",
            "The circumcenter is where the three PERPENDICULAR BISECTORS of the sides meet. Those are perpendicular to the sides like altitudes are, but they start at the midpoints rather than at the opposite vertices.",
            "Correct. An altitude runs from a vertex perpendicular to the opposite side, and all three are concurrent at the orthocenter. In an obtuse triangle two of the altitudes have to be extended outside the figure to meet, so the orthocenter itself lands outside the triangle."
          ],
          "hint": "Match the line to the center: altitudes, medians, angle bisectors, and perpendicular bisectors each produce a different point of concurrency.",
          "diagram": "<svg viewBox=\"0 0 240 140\" width=\"100%\" style=\"max-width:240px; height:auto; display:block; margin:0 auto;\" role=\"img\"><polygon points=\"35,120 205,120 140,25\" fill=\"none\" stroke=\"#7c3aed\" stroke-width=\"2\" stroke-linejoin=\"round\"/><line x1=\"140\" y1=\"25\" x2=\"140\" y2=\"120\" stroke=\"#1f2024\" stroke-width=\"1.6\" stroke-linecap=\"round\"/><line x1=\"35\" y1=\"120\" x2=\"175\" y2=\"65\" stroke=\"#1f2024\" stroke-width=\"1.6\" stroke-linecap=\"round\"/><line x1=\"205\" y1=\"120\" x2=\"96\" y2=\"62\" stroke=\"#1f2024\" stroke-width=\"1.6\" stroke-linecap=\"round\"/><path d=\"M 130,120 L 130,110 L 140,110\" fill=\"none\" stroke=\"#6b6b76\" stroke-width=\"1.4\"/><circle cx=\"140\" cy=\"82\" r=\"3.2\" fill=\"#c0392b\"/><text x=\"146\" y=\"80\" fill=\"#c0392b\" font-size=\"10.5\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">orthocenter</text></svg>",
          "diagramCaption": "The three altitudes are concurrent. Not drawn to scale"
        },
        {
          "id": "tct-002",
          "topic": "Triangle Centers",
          "difficulty": "medium",
          "stem": "Which triangle center is ALWAYS equidistant from the three vertices of a triangle?",
          "choices": [
            "Orthocenter",
            "Circumcenter",
            "Centroid",
            "Incenter"
          ],
          "correctIndex": 1,
          "explanations": [
            "The orthocenter is the concurrency of the altitudes. It has no equidistance property at all — in an obtuse triangle it is not even inside the figure.",
            "Correct. The circumcenter is the center of the circumscribed circle, the one passing through all three vertices. Every vertex sits on that circle, so every vertex is exactly one radius away. It is where the perpendicular bisectors meet, and a perpendicular bisector is precisely the set of points equidistant from two endpoints.",
            "The centroid is the balance point, dividing each median 2:1. It is equidistant from the vertices only when the triangle is equilateral.",
            "The incenter is equidistant from the three SIDES, not the vertices — it is the center of the inscribed circle, which touches each side rather than passing through each corner."
          ],
          "hint": "One of these four centers is the center of a circle. Ask which circle — the one through the corners, or the one tucked inside touching the edges?",
          "diagram": "<svg viewBox=\"0 0 240 140\" width=\"100%\" style=\"max-width:240px; height:auto; display:block; margin:0 auto;\" role=\"img\"><polygon points=\"35,120 205,120 140,25\" fill=\"none\" stroke=\"#7c3aed\" stroke-width=\"2\" stroke-linejoin=\"round\"/><circle cx=\"120\" cy=\"82\" r=\"88\" fill=\"none\" stroke=\"#6b6b76\" stroke-width=\"1.5\"/><circle cx=\"120\" cy=\"82\" r=\"3.2\" fill=\"#c0392b\"/><line x1=\"120\" y1=\"82\" x2=\"35\" y2=\"120\" stroke=\"#c0392b\" stroke-width=\"1.5\" stroke-dasharray=\"5 4\" stroke-linecap=\"round\"/><line x1=\"120\" y1=\"82\" x2=\"205\" y2=\"120\" stroke=\"#c0392b\" stroke-width=\"1.5\" stroke-dasharray=\"5 4\" stroke-linecap=\"round\"/><line x1=\"120\" y1=\"82\" x2=\"140\" y2=\"25\" stroke=\"#c0392b\" stroke-width=\"1.5\" stroke-dasharray=\"5 4\" stroke-linecap=\"round\"/><text x=\"126\" y=\"78\" fill=\"#c0392b\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">R</text></svg>",
          "diagramCaption": "The circumscribed circle passes through all three vertices. Not drawn to scale"
        },
        {
          "id": "tct-003",
          "topic": "Triangle Centers",
          "difficulty": "medium",
          "stem": "Which triangle center is ALWAYS equidistant from the three SIDES of a triangle?",
          "choices": [
            "Centroid",
            "Orthocenter",
            "Circumcenter",
            "Incenter"
          ],
          "correctIndex": 3,
          "explanations": [
            "The centroid is the balance point of the triangle, found from the medians. Its distances to the three sides are generally all different.",
            "The orthocenter is where the altitudes meet. Altitudes measure perpendicular distance from a VERTEX to a side, which is a different quantity from the distance from the center to each side — and the orthocenter has no equidistance property.",
            "The circumcenter is equidistant from the three VERTICES, not the sides. Swapping these two is the most common error on this question.",
            "Correct. The incenter is the center of the inscribed circle — the incircle — which is tangent to all three sides, so the perpendicular distance from the incenter to each side is one radius. It is where the angle bisectors meet, and an angle bisector is the set of points equidistant from the two sides forming the angle."
          ],
          "hint": "The inscribed circle touches each side once. What must be true of the distance from its center to each side?",
          "diagram": "<svg viewBox=\"0 0 240 140\" width=\"100%\" style=\"max-width:240px; height:auto; display:block; margin:0 auto;\" role=\"img\"><polygon points=\"35,120 205,120 140,25\" fill=\"none\" stroke=\"#7c3aed\" stroke-width=\"2\" stroke-linejoin=\"round\"/><circle cx=\"128\" cy=\"87\" r=\"30\" fill=\"none\" stroke=\"#6b6b76\" stroke-width=\"1.5\"/><circle cx=\"128\" cy=\"87\" r=\"3.2\" fill=\"#c0392b\"/><line x1=\"128\" y1=\"87\" x2=\"128\" y2=\"120\" stroke=\"#c0392b\" stroke-width=\"1.5\" stroke-dasharray=\"5 4\" stroke-linecap=\"round\"/><text x=\"134\" y=\"84\" fill=\"#c0392b\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">r</text></svg>",
          "diagramCaption": "The inscribed circle is tangent to all three sides. Not drawn to scale"
        },
        {
          "id": "tct-004",
          "topic": "Triangle Centers",
          "difficulty": "medium",
          "stem": "The centroid of a triangle divides each median into two segments. What is the ratio of the vertex-to-centroid segment to the centroid-to-midpoint segment?",
          "choices": [
            "3 : 2",
            "2 : 1",
            "1 : 1",
            "3 : 1"
          ],
          "correctIndex": 1,
          "explanations": [
            "3 : 2 does not describe either split. It may be a half-memory of the fact that the vertex-to-centroid piece is 2/3 of the whole median — but 2/3 to 1/3 reduces to 2 : 1, not 3 : 2.",
            "Correct. The centroid splits every median so that the piece connecting to the vertex is twice as long as the piece connecting to the side midpoint — a 2 : 1 ratio. Equivalently, the centroid sits 2/3 of the way along each median from its vertex.",
            "1 : 1 would mean the centroid is the midpoint of each median. It is not; that would make it the balance point of the three midpoints rather than of the triangle's area.",
            "3 : 1 would put the centroid 3/4 of the way from the vertex. The correct fraction is 2/3."
          ],
          "hint": "The centroid sits two-thirds of the way from each vertex. Turn that fraction into a ratio of the two pieces.",
          "diagram": "<svg viewBox=\"0 0 240 140\" width=\"100%\" style=\"max-width:240px; height:auto; display:block; margin:0 auto;\" role=\"img\"><polygon points=\"30,120 210,120 120,25\" fill=\"none\" stroke=\"#7c3aed\" stroke-width=\"2\" stroke-linejoin=\"round\"/><line x1=\"120\" y1=\"25\" x2=\"120\" y2=\"120\" stroke=\"#1f2024\" stroke-width=\"1.8\" stroke-linecap=\"round\"/><circle cx=\"120\" cy=\"87\" r=\"3.2\" fill=\"#c0392b\"/><circle cx=\"120\" cy=\"120\" r=\"3.2\" fill=\"#6b6b76\"/><text x=\"126\" y=\"60\" fill=\"#1f2024\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">2 parts</text><text x=\"126\" y=\"110\" fill=\"#1f2024\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">1 part</text><text x=\"60\" y=\"118\" fill=\"#6b6b76\" font-size=\"10\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">midpoint</text></svg>",
          "diagramCaption": "The centroid divides the median 2 : 1 from the vertex. Not drawn to scale"
        },
        {
          "id": "tct-005",
          "topic": "The Euler Line",
          "difficulty": "hard",
          "stem": "Leonhard Euler proved that three triangle centers always lie on a single line (the Euler Line). Which center is NOT on the Euler Line?",
          "choices": [
            "Circumcenter",
            "Orthocenter",
            "Incenter",
            "Centroid"
          ],
          "correctIndex": 2,
          "explanations": [
            "The circumcenter is one of the three that ARE collinear. It sits at one end of the Euler Line.",
            "The orthocenter is on the Euler Line, at the opposite end from the circumcenter.",
            "Correct. The incenter is the odd one out. The circumcenter, centroid and orthocenter are always collinear — and the centroid always lies between the other two, twice as far from the orthocenter as from the circumcenter. The incenter joins them only when the triangle is isosceles or equilateral, where symmetry forces every center onto the axis.",
            "The centroid is on the Euler Line, and in a specific place: exactly one third of the way from the circumcenter to the orthocenter."
          ],
          "hint": "Three of these four are built from perpendicularity or from midpoints. The fourth is built from angles — and that is the one that breaks ranks.",
          "diagram": "<svg viewBox=\"0 0 240 136\" width=\"100%\" style=\"max-width:240px; height:auto; display:block; margin:0 auto;\" role=\"img\"><polygon points=\"30,118 212,118 92,28\" fill=\"none\" stroke=\"#7c3aed\" stroke-width=\"2\" stroke-linejoin=\"round\"/><line x1=\"24\" y1=\"96\" x2=\"220\" y2=\"60\" stroke=\"#c0392b\" stroke-width=\"1.6\" stroke-linecap=\"round\"/><circle cx=\"58\" cy=\"90\" r=\"3.2\" fill=\"#c0392b\"/><circle cx=\"112\" cy=\"80\" r=\"3.2\" fill=\"#c0392b\"/><circle cx=\"166\" cy=\"70\" r=\"3.2\" fill=\"#c0392b\"/><text x=\"30\" y=\"84\" fill=\"#c0392b\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">O</text><text x=\"108\" y=\"74\" fill=\"#c0392b\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">G</text><text x=\"170\" y=\"64\" fill=\"#c0392b\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">H</text><circle cx=\"96\" cy=\"96\" r=\"3.2\" fill=\"#1f2024\"/><text x=\"78\" y=\"108\" fill=\"#1f2024\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">I</text></svg>",
          "diagramCaption": "O, G and H are collinear; I sits off the line. Not drawn to scale"
        },
        {
          "id": "tct-006",
          "topic": "Triangle Centers",
          "difficulty": "easy",
          "stem": "Which set of lines intersects at the circumcenter of a triangle?",
          "choices": [
            "Altitudes from vertices to opposite sides",
            "Medians from vertices to midpoints",
            "Perpendicular bisectors of the sides",
            "Angle bisectors of the vertices"
          ],
          "correctIndex": 2,
          "explanations": [
            "Altitudes meet at the orthocenter. They are perpendicular to the sides, like perpendicular bisectors, but they pass through the opposite vertex rather than through the side's midpoint.",
            "Medians meet at the centroid. They pass through midpoints, like perpendicular bisectors, but they run to the opposite vertex rather than perpendicular to the side.",
            "Correct. The perpendicular bisectors of the three sides are concurrent at the circumcenter. Each one is the locus of points equidistant from the two endpoints of its side, so their common point is equidistant from all three vertices — which is exactly the center of the circumscribed circle.",
            "Angle bisectors meet at the incenter."
          ],
          "hint": "The perpendicular bisector of a segment is the set of points equidistant from its endpoints. What does intersecting three of them guarantee?"
        },
        {
          "id": "tct-007",
          "topic": "Triangle Centers",
          "difficulty": "hard",
          "stem": "In an obtuse triangle, where do the circumcenter and orthocenter lie?",
          "choices": [
            "Both inside the triangle",
            "Both outside the triangle",
            "The circumcenter inside, the orthocenter outside",
            "Both on the longest side"
          ],
          "correctIndex": 1,
          "explanations": [
            "Both are inside only when the triangle is acute. The obtuse case is exactly where that stops being true, which is why it gets asked.",
            "Correct. In an obtuse triangle both fall outside. The circumcenter moves across the longest side because the circumscribed circle's center must sit on the far side of it, and the orthocenter moves out because two of the three altitudes only meet the opposite sides once those sides are extended. The centroid, by contrast, is always inside — it is an average of the vertices.",
            "This mixes the two up, and neither half is right for an obtuse triangle. Both leave the figure together.",
            "Only in a right triangle does a center land on a side: there the circumcenter is the midpoint of the hypotenuse, and the orthocenter is the right-angle vertex."
          ],
          "hint": "Think about the right-triangle case first — the circumcenter sits on the hypotenuse. Now push the angle past 90° and ask which way it keeps moving."
        },
        {
          "id": "tct-008",
          "topic": "Midsegments",
          "difficulty": "medium",
          "stem": "A midsegment of a triangle joins the midpoints of two sides. If the third side measures 18 cm, what is the length of that midsegment, and how is it oriented?",
          "choices": [
            "9 cm, parallel to the third side",
            "9 cm, perpendicular to the third side",
            "36 cm, parallel to the third side",
            "18 cm, parallel to the third side"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. The Triangle Midsegment Theorem gives both facts at once: the midsegment is parallel to the third side and exactly half its length. Half of 18 cm is 9 cm.",
            "The length is right but the orientation is not. Perpendicularity is not part of the theorem, and the midsegment is generally not perpendicular to anything.",
            "36 cm doubles instead of halving. The midsegment is the shorter of the two — it spans a smaller, similar triangle cut off at the top.",
            "18 cm would make the midsegment congruent to the third side. That happens in a parallelogram, not here: the midsegment and the third side are two parallel sides of a trapezoid whose lengths differ by a factor of two."
          ],
          "hint": "The midsegment cuts off a small triangle similar to the original, with a scale factor of one half. That factor answers both parts.",
          "diagram": "<svg viewBox=\"0 0 240 145\" width=\"100%\" style=\"max-width:240px; height:auto; display:block; margin:0 auto;\" role=\"img\"><polygon points=\"30,120 210,120 110,25\" fill=\"none\" stroke=\"#7c3aed\" stroke-width=\"2\" stroke-linejoin=\"round\"/><line x1=\"70\" y1=\"72\" x2=\"160\" y2=\"72\" stroke=\"#1f2024\" stroke-width=\"2.2\" stroke-linecap=\"round\"/><circle cx=\"70\" cy=\"72\" r=\"3.2\" fill=\"#1f2024\"/><circle cx=\"160\" cy=\"72\" r=\"3.2\" fill=\"#1f2024\"/><text x=\"98\" y=\"66\" fill=\"#1f2024\" font-size=\"10.5\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">midsegment</text><text x=\"104\" y=\"136\" fill=\"#6b6b76\" font-size=\"11\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">18 cm</text></svg>",
          "diagramCaption": "The midsegment joins two side midpoints. Not drawn to scale"
        },
        {
          "id": "tct-009",
          "topic": "Right Triangle Altitudes",
          "difficulty": "hard",
          "stem": "An altitude drawn to the hypotenuse of a right triangle splits it into segments of length 4 and 9. What is the length of the altitude?",
          "choices": [
            "6",
            "6.5",
            "13",
            "36"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. The altitude to the hypotenuse is the geometric mean of the two segments it creates: h = √(a·b) = √(4 × 9) = √36 = 6. This follows from the two smaller triangles being similar to each other, which gives h/4 = 9/h, so h² = 36.",
            "6.5 is the arithmetic mean, (4 + 9)/2. The relationship here is a GEOMETRIC mean — multiply and take the root, don't add and halve.",
            "13 is the whole hypotenuse, 4 + 9. That is the segment the altitude divides, not the altitude itself.",
            "36 is h², one step short. The equation h² = 4 × 9 still needs a square root taken."
          ],
          "hint": "The altitude creates two smaller triangles, each similar to the whole. Set up the proportion h/4 = 9/h and see what it gives.",
          "diagram": "<svg viewBox=\"0 0 240 136\" width=\"100%\" style=\"max-width:240px; height:auto; display:block; margin:0 auto;\" role=\"img\"><polygon points=\"25,110 215,110 77,25\" fill=\"none\" stroke=\"#7c3aed\" stroke-width=\"2\" stroke-linejoin=\"round\"/><line x1=\"77\" y1=\"25\" x2=\"77\" y2=\"110\" stroke=\"#1f2024\" stroke-width=\"2\" stroke-linecap=\"round\"/><path d=\"M 77,98 L 89,98 L 89,110\" fill=\"none\" stroke=\"#6b6b76\" stroke-width=\"1.4\"/><path d=\"M 68,34 L 78,40 L 72,48\" fill=\"none\" stroke=\"#6b6b76\" stroke-width=\"1.4\"/><text x=\"50\" y=\"126\" fill=\"#6b6b76\" font-size=\"12\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">4</text><text x=\"140\" y=\"126\" fill=\"#6b6b76\" font-size=\"12\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">9</text><text x=\"82\" y=\"72\" fill=\"#1f2024\" font-size=\"13\" font-family=\"system-ui, sans-serif\" font-weight=\"600\" paint-order=\"stroke\" stroke=\"#ffffff\" stroke-width=\"3.5\" stroke-linejoin=\"round\">h</text></svg>",
          "diagramCaption": "The altitude falls to the hypotenuse. Not drawn to scale"
        },
        {
          "id": "tct-010",
          "topic": "Triangle Inequality",
          "difficulty": "medium",
          "stem": "Two sides of a triangle measure 7 and 10. What is the full range of possible lengths for the third side?",
          "choices": [
            "Greater than 3 and less than 17",
            "Greater than 7 and less than 10",
            "Greater than 0 and less than 17",
            "Greater than 3 and less than 10"
          ],
          "correctIndex": 0,
          "explanations": [
            "Correct. The Triangle Inequality has to hold all three ways. The third side must be less than 7 + 10 = 17, and greater than 10 − 7 = 3 (otherwise the two given sides could not reach across it). So 3 < x < 17, with both bounds strict.",
            "7 to 10 confuses the range of the third side with the two given lengths themselves. Nothing requires the third side to fall between them — a triangle with sides 7, 10, 15 is perfectly valid.",
            "Getting the upper bound right but dropping the lower one misses half the constraint. A third side of 1 fails: 7 + 1 = 8, which is less than 10, so those three lengths cannot close.",
            "The lower bound is right but the upper is not. 10 is the length of a given side, not a cap — the sum 7 + 10 = 17 is what caps it."
          ],
          "hint": "Two conditions, not one: the third side must be short enough for the other two to reach across, and long enough that they cannot swallow it."
        },
        {
          "id": "tct-011",
          "topic": "Triangle Centers",
          "difficulty": "medium",
          "stem": "In which type of triangle do the centroid, incenter, circumcenter and orthocenter all coincide at a single point?",
          "choices": [
            "Any isosceles triangle",
            "Any right triangle",
            "An equilateral triangle",
            "Any obtuse triangle"
          ],
          "correctIndex": 2,
          "explanations": [
            "An isosceles triangle has one axis of symmetry, so all four centers land ON that axis — which is why the incenter joins the Euler Line there. But they are still four distinct points spread along it unless the third side matches too.",
            "In a right triangle the centers are conspicuously spread out: the circumcenter sits at the hypotenuse's midpoint and the orthocenter at the right-angle vertex, which are different points.",
            "Correct. An equilateral triangle has three axes of symmetry, and every center must lie on all of them. Three distinct lines meet in at most one point, so all four centers collapse together. Each cevian is simultaneously a median, an altitude, an angle bisector and a perpendicular bisector.",
            "Obtuse triangles push the circumcenter and orthocenter outside the figure entirely, while the centroid stays inside — about as far from coinciding as they get."
          ],
          "hint": "Each center must sit on every axis of symmetry the triangle has. How many axes force them all onto one point?"
        },
        {
          "id": "tct-012",
          "topic": "The Euler Line",
          "difficulty": "hard",
          "stem": "On the Euler Line, the centroid G lies between the circumcenter O and the orthocenter H. What is the ratio OG : GH?",
          "choices": [
            "1 : 1",
            "1 : 2",
            "2 : 1",
            "1 : 3"
          ],
          "correctIndex": 1,
          "explanations": [
            "1 : 1 would make the centroid the midpoint of OH. It is not — the centroid sits noticeably closer to the circumcenter.",
            "Correct. OG : GH = 1 : 2. The centroid is one third of the way from the circumcenter to the orthocenter, so OH = 3·OG. The same 2 : 1 proportion that governs the medians reappears here, which is not a coincidence: both come from the homothety centered at G with ratio −2 that carries the medial triangle onto the original.",
            "2 : 1 reverses it, putting the centroid closer to the orthocenter. The distances are the other way round.",
            "1 : 3 would make OH = 4·OG. The correct factor is 3."
          ],
          "hint": "The centroid is one third of the way along OH. Turn that fraction into the ratio of the two pieces."
        },
        {
          "id": "tct-013",
          "topic": "Centroid Ratio",
          "difficulty": "medium",
          "stem": "The centroid of a triangle divides a median so that the longer piece measures 10. What is the length of the entire median?",
          "choices": [
            "12",
            "15",
            "20",
            "30"
          ],
          "correctIndex": 1,
          "explanations": [
            "12 would follow from a 5:1 split. The centroid splits each median in a 2:1 ratio measured from the vertex, so the two pieces are 10 and 5.",
            "Correct. The centroid divides every median in a 2:1 ratio, with the longer piece adjacent to the vertex. If the longer piece is 10, the shorter is 5, and the full median is 10 + 5 = 15.",
            "20 doubles the longer piece, which would be right only if the centroid were the MIDPOINT of the median. It is not — the midpoint of a median is not a triangle center at all.",
            "30 triples the longer piece. The whole median is 3 halves of the shorter piece, not 3 times the longer one — the ratio applies to the parts, not to the total directly."
          ],
          "hint": "The 2:1 ratio describes the two pieces relative to each other. Find the shorter piece first, then add."
        },
        {
          "id": "tct-014",
          "topic": "Incenter",
          "difficulty": "easy",
          "stem": "Which triangle center is always equidistant from all three SIDES of the triangle?",
          "choices": [
            "Centroid",
            "Circumcenter",
            "Incenter",
            "Orthocenter"
          ],
          "correctIndex": 2,
          "explanations": [
            "The centroid is the balance point, where the medians meet. It is the center of mass, but its distances to the three sides are generally all different.",
            "The circumcenter is equidistant from the three VERTICES, not the sides — that shared distance is the circumradius. Vertices and sides are the distinction the question turns on.",
            "Correct. The incenter is where the three angle bisectors meet, and a point on an angle bisector is equidistant from the two sides forming that angle. Meeting all three bisectors makes it equidistant from all three sides, which is exactly the radius of the inscribed circle.",
            "The orthocenter is where the altitudes meet. It carries no equidistance property at all, and in an obtuse triangle it falls outside the triangle entirely."
          ],
          "hint": "Two of these centers are equidistant from something. Sort out which one measures to sides and which to vertices."
        },
        {
          "id": "tct-015",
          "topic": "Circumcenter Location",
          "difficulty": "hard",
          "stem": "In which type of triangle does the circumcenter lie OUTSIDE the triangle?",
          "choices": [
            "Acute",
            "Right",
            "Obtuse",
            "Equilateral"
          ],
          "correctIndex": 2,
          "explanations": [
            "In an acute triangle the circumcenter lies strictly inside. All three angles being under 90° keeps the perpendicular bisectors meeting within the figure.",
            "In a right triangle the circumcenter lies exactly ON the triangle — at the midpoint of the hypotenuse. That is a boundary case rather than an outside one, and it is why the hypotenuse is a diameter of the circumscribed circle.",
            "Correct. In an obtuse triangle the circumcenter falls outside, on the far side of the longest side from the obtuse angle. The circumscribed circle still passes through all three vertices; its center simply is not enclosed by the triangle.",
            "An equilateral triangle is a special acute triangle in which the circumcenter, incenter, centroid and orthocenter all coincide at a single interior point."
          ],
          "hint": "Work through the three angle classifications in order, and remember the right-triangle case lands somewhere very specific."
        },
        {
          "id": "tct-016",
          "topic": "The Euler Line",
          "difficulty": "hard",
          "stem": "Which three triangle centers are always collinear, lying on the Euler line?",
          "choices": [
            "Incenter, centroid, circumcenter",
            "Centroid, circumcenter, orthocenter",
            "Incenter, orthocenter, circumcenter",
            "All four classical centers"
          ],
          "correctIndex": 1,
          "explanations": [
            "The incenter is the odd one out — it does NOT lie on the Euler line in a general triangle. Swapping it in for the orthocenter is the usual way this question is missed.",
            "Correct. The centroid, circumcenter and orthocenter are always collinear, and the line through them is the Euler line. The centroid sits between the other two and divides the segment in a 2:1 ratio, with the longer part toward the orthocenter.",
            "This again includes the incenter, which is not on the Euler line except in special triangles such as isosceles ones, where symmetry forces every center onto the axis.",
            "All four coincide only in an equilateral triangle. In that case the Euler line degenerates to a single point, so it is a special case rather than the general statement."
          ],
          "hint": "One of the four classical centers is famously absent from this line. Identify which, and the rest follows."
        },
        {
          "id": "tct-017",
          "topic": "Orthocenter",
          "difficulty": "medium",
          "stem": "A triangle has a 120° angle. Where is its orthocenter located?",
          "choices": [
            "Inside the triangle",
            "On the triangle, at a vertex",
            "Outside the triangle",
            "At the midpoint of the longest side"
          ],
          "correctIndex": 2,
          "explanations": [
            "The orthocenter lies inside only when all three angles are acute. A 120° angle makes this triangle obtuse, so the interior case does not apply.",
            "The orthocenter sits at a vertex only in a RIGHT triangle, where it lands on the right-angle vertex because two of the altitudes are the legs themselves. A 120° angle is obtuse, not right.",
            "Correct. In an obtuse triangle two of the altitudes must be extended beyond the triangle before they meet, so their point of concurrency falls outside the figure. The 120° angle guarantees this.",
            "The midpoint of the longest side is where the CIRCUMCENTER lands in a right triangle. It is not a location the orthocenter takes, and this triangle is obtuse in any case."
          ],
          "hint": "Classify the triangle by its largest angle first, then recall how the altitudes behave in that case."
        },
        {
          "id": "tct-018",
          "topic": "Reasoning: Choose the Centre",
          "difficulty": "hard",
          "stem": "A town wants to place a water tower equally distant from three straight roads that form a triangle. Which centre locates it, and why?",
          "choices": [
            "The circumcentre, because it is equidistant from the three vertices.",
            "The incentre, because it is equidistant from the three SIDES.",
            "The centroid, because it balances the triangle.",
            "The orthocentre, because the altitudes measure distance to the sides."
          ],
          "correctIndex": 1,
          "explanations": [
            "The circumcentre is equidistant from the three VERTICES — the right choice for reaching three towns, not three roads.",
            "Correct. The incentre is where the angle bisectors meet, and it is equidistant from the three sides. Roads are the sides, so this is the point that is equally far from each road.",
            "The centroid is the balance point, the intersection of the medians. It is generally not equidistant from anything in particular.",
            "The orthocentre is where the altitudes meet. Altitudes are perpendicular to the sides, but their intersection is not equidistant from them."
          ],
          "hint": "Distance to a LINE is measured perpendicular. Which centre is defined by equal perpendicular distances?",
          "kind": "reasoning"
        },
        {
          "id": "tct-019",
          "topic": "Reasoning: Error Analysis",
          "difficulty": "medium",
          "stem": "A student claims every triangle centre always lies inside the triangle. Which centre disproves this, and when?",
          "choices": [
            "The centroid, when the triangle is scalene.",
            "The incentre, when the triangle is obtuse.",
            "The circumcentre, which falls outside whenever the triangle is obtuse.",
            "None — all four centres are always interior."
          ],
          "correctIndex": 2,
          "explanations": [
            "The centroid is always interior: it is the average of the three vertices, so it can never leave the triangle.",
            "The incentre is always interior too — it is the centre of a circle that fits inside the triangle.",
            "Correct. The circumcentre lies outside an obtuse triangle (and exactly on the hypotenuse's midpoint for a right triangle). The orthocentre also leaves an obtuse triangle.",
            "Two of them can be exterior. Only the centroid and incentre are always inside."
          ],
          "kind": "reasoning"
        }
      ]
    }
  }
}