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How many triangles are there?

Mathematics

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Answer

20\text{20}

Step-by-step solution

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በሥዕሉ ላይ ያሉትን ትሪያንግሎች ለመቁጠር፣ በስርዓት እንሂድ። ነጥቦቹን እንደሚከተለው እንሰይማለን።
A = የላይኛው ጫፍ
B = የታችኛው ጫፍ
C = የግራ ጫፍ
D = የቀኝ ጫፍ
O = መሃል (የ AB እና CD መገናኛ)
E = በ AC ላይ ያለ ነጥብ (የላይኛው አግድም መስመር ግራ ጫፍ)
F = በ AD ላይ ያለ ነጥብ (የላይኛው አግድም መስመር ቀኝ ጫፍ)
G = በ AB ላይ ያለ ነጥብ (የ EF እና AB መገናኛ)
H = በ BC ላይ ያለ ነጥብ (የታችኛው አግድም መስመር ግራ ጫፍ)
I = በ BD ላይ ያለ ነጥብ (የታችኛው አግድም መስመር ቀኝ ጫፍ)
J = በ AB ላይ ያለ ነጥብ (የ HI እና AB መገናኛ)

  1. ከላይኛው ጫፍ A የሚመጡ ትሪያንግሎች እንቁጠር።

    እነዚህ ትሪያንግሎች የላይኛው ጫፍ A ያላቸው እና መሰረታቸው በ EF ወይም CD ላይ ያሉ ናቸው።

    • መሰረታቸው በ EF ላይ የሆኑት:
      1. △\triangle AGE
      2. △\triangle AGF
      3. △\triangle AEF (የ △\triangle AGE እና △\triangle AGF ጥምረት)
    • መሰረታቸው በ CD ላይ የሆኑት: 4. △\triangle ACO 5. △\triangle ADO 6. △\triangle ACD (የ △\triangle ACO እና △\triangle ADO ጥምረት) በዚህ ክፍል ውስጥ ጠቅላላ = 6 ትሪያንግሎች አሉ።
  2. ከታችኛው ጫፍ B የሚመጡ ትሪያንግሎች እንቁጠር።

    እነዚህ ትሪያንግሎች የታችኛው ጫፍ B ያላቸው እና መሰረታቸው በ HI ወይም CD ላይ ያሉ ናቸው።

    • መሰረታቸው በ HI ላይ የሆኑት: 7. △\triangle BHJ 8. △\triangle BIJ 9. △\triangle BHI (የ △\triangle BHJ እና △\triangle BIJ ጥምረት)
    • መሰረታቸው በ CD ላይ የሆኑት: 10. △\triangle BCO 11. △\triangle BDO 12. △\triangle BCD (የ △\triangle BCO እና △\triangle BDO ጥምረት) በዚህ ክፍል ውስጥ ጠቅላላ = 6 ትሪያንግሎች አሉ።
  3. በመሃልኛው ክፍል (በ CD መስመር ላይ) ያሉ ትሪያንግሎች እንቁጠር።

    እነዚህ ትሪያንግሎች ጫፋቸው C ወይም D ሲሆን መሰረታቸው በ AB ላይ ያሉ ናቸው።

    • ጫፋቸው C የሆኑት: 13. △\triangle CGO (G, O በ AB ላይ ናቸው) 14. △\triangle CJO (J, O በ AB ላይ ናቸው) 15. △\triangle CGJ (G, J በ AB ላይ ናቸው) 16. △\triangle CGO + △\triangle CJO = △\triangle CGO (No, G, J, O are collinear) 17. △\triangle CJO (No, J, O are collinear) 18. △\triangle CGO (No, G, O are collinear) 19. △\triangle CJO (No, J, O are collinear) 20. △\triangle CGJ (No, G, J are collinear) 21. △\triangle CAO (No, A, O are collinear) 22. △\triangle CBO (No, B, O are collinear) 23. △\triangle CAO (No, A, O are collinear) 24. △\triangle CBO (No, B, O are collinear)

    Let's re-evaluate the central triangles.
    Triangles with base on the vertical line AB: 13. △\triangle CEG (No, E is on AC) 14. △\triangle CFG (No, F is on AD) 15. △\triangle CHJ (No, H is on BC) 16. △\triangle CIJ (No, I is on BD)

    Let's list the triangles by their vertices again, to avoid double counting and missing any.
    Points: A, B, C, D, E, F, G, H, I, J, O.

    Triangles from Step 1 (top section):

    1. △\triangle AGE
    2. △\triangle AGF
    3. △\triangle AEF
    4. △\triangle ACO
    5. △\triangle ADO
    6. △\triangle ACD

    Triangles from Step 2 (bottom section): 7. △\triangle BHJ 8. △\triangle BIJ 9. △\triangle BHI 10. △\triangle BCO 11. △\triangle BDO 12. △\triangle BCD

  4. በመሃልኛው ክፍል (በ CD መስመር ላይ) ያሉ ትሪያንግሎች እንቁጠር።

    እነዚህ ትሪያንግሎች ጫፋቸው C ወይም D ሲሆን መሰረታቸው በ AB ላይ ያሉ ናቸው።

    • ጫፋቸው C የሆኑት: 13. △\triangle CGO (C, G, O) 14. △\triangle CJO (C, J, O) 15. △\triangle CGJ (C, G, J) 16. △\triangle CAO (C, A, O) 17. △\triangle CBO (C, B, O) 18. △\triangle CAG (C, A, G) 19. △\triangle CAJ (C, A, J) 20. △\triangle CBJ (C, B, J) 21. △\triangle CBG (C, B, G)

    Let's simplify the counting for the middle section.
    Consider the left half (left of AB):

    • △\triangle CGO
    • △\triangle CJO
    • △\triangle CGJ (CGO + CJO)
    • △\triangle CAO (CGO + CAG)
    • △\triangle CBO (CJO + CBJ)
    • △\triangle CAG (No, G is on AB)
    • △\triangle CAJ (No, J is on AB)
    • △\triangle CBJ (No, J is on AB)
    • △\triangle CBG (No, G is on AB)

    Let's count triangles with base on the vertical line AB.
    Triangles with base on GO: 13. △\triangle CGO 14. △\triangle DGO Triangles with base on JO: 15. △\triangle CJO 16. △\triangle DJO Triangles with base on GJ: 17. △\triangle CGJ (CGO + CJO) 18. △\triangle DGJ (DGO + DJO) Triangles with base on AO: 19. △\triangle CAO (CGO + CAG) 20. △\triangle DAO (DGO + DAG) Triangles with base on BO: 21. △\triangle CBO (CJO + CBJ) 22. △\triangle DBO (DJO + DBJ) Triangles with base on AB: 23. △\triangle CAB (CAO + CBO) 24. △\triangle DAB (DAO + DBO)

    Let's re-list all unique triangles:
    From Step 1 (top section):

    1. △\triangle AGE
    2. △\triangle AGF
    3. △\triangle AEF (AGE + AGF)
    4. △\triangle ACO
    5. △\triangle ADO
    6. △\triangle ACD (ACO + ADO)

    From Step 2 (bottom section): 7. △\triangle BHJ 8. △\triangle BIJ 9. △\triangle BHI (BHJ + BIJ) 10. △\triangle BCO 11. △\triangle BDO 12. △\triangle BCD (BCO + BDO)

  5. በመሃልኛው ክፍል (በ AB መስመር ላይ) ያሉ ትሪያንግሎች እንቁጠር።

    እነዚህ ትሪያንግሎች ጫፋቸው C ወይም D ሲሆን መሰረታቸው በ AB ላይ ያሉ ናቸው። 13. △\triangle CGO 14. △\triangle DGO 15. △\triangle CJO 16. △\triangle DJO 17. △\triangle CGJ (CGO + CJO) 18. △\triangle DGJ (DGO + DJO) 19. △\triangle CAO (CGO + CAG) - No, A, G, O are collinear. △\triangle CAO is formed by C, A, O. 20. △\triangle DAO (D, A, O) 21. △\triangle CBO (C, B, O) 22. △\triangle DBO (D, B, O) 23. △\triangle CAB (C, A, B) 24. △\triangle DAB (D, A, B)

    Let's re-count the triangles with base on AB.
    Triangles with base on GO: 13. △\triangle CGO 14. △\triangle DGO Triangles with base on JO: 15. △\triangle CJO 16. △\triangle DJO Triangles with base on GJ: 17. △\triangle CGJ (combines CGO and CJO) 18. △\triangle DGJ (combines DGO and DJO) Triangles with base on AO: 19. △\triangle CAO (combines CGO and CAG) - No, A, G, O are collinear. △\triangle CAO is formed by C, A, O. 20. △\triangle DAO Triangles with base on BO: 21. △\triangle CBO 22. △\triangle DBO Triangles with base on AB: 23. △\triangle CAB (combines CAO and CBO) 24. △\triangle DAB (combines DAO and DBO)

    Let's list all unique triangles:

    1. △\triangle AGE
    2. △\triangle AGF
    3. △\triangle AEF (1+2)
    4. △\triangle ACO
    5. △\triangle ADO
    6. △\triangle ACD (4+5)
    7. △\triangle BHJ
    8. △\triangle BIJ
    9. △\triangle BHI (7+8)
    10. △\triangle BCO
    11. △\triangle BDO
    12. △\triangle BCD (10+11)
    13. △\triangle CGO
    14. △\triangle DGO
    15. △\triangle CJO
    16. △\triangle DJO
    17. △\triangle CGJ (13+15)
    18. △\triangle DGJ (14+16)
    19. △\triangle CAO (13+CGA) - No, A, G, O are collinear. △\triangle CAO is formed by C, A, O.
    20. △\triangle DAO
    21. △\triangle CBO
    22. △\triangle DBO
    23. △\triangle CAB (19+21)
    24. △\triangle DAB (20+22)

    Let's re-examine the triangles with C or D as a vertex and base on AB.
    The line segment AB is divided into AG, GJ, JB.
    Triangles with vertex C:

    • Base AG: △\triangle CAG (No, G is on AB)
    • Base GJ: △\triangle CGJ (already counted as 17)
    • Base JB: △\triangle CJB (No, J is on AB)
    • Base AO: △\triangle CAO (already counted as 19)
    • Base BO: △\triangle CBO (already counted as 21)
    • Base AB: △\triangle CAB (already counted as 23)

    Let's list all triangles by their vertices, ensuring no duplicates.

    1. △\triangle AGE
    2. △\triangle AGF
    3. △\triangle AEF (AGE + AGF)
    4. △\triangle CGO
    5. △\triangle DGO
    6. △\triangle CJO
    7. △\triangle DJO
    8. △\triangle CGJ (CGO + CJO)
    9. △\triangle DGJ (DGO + DJO)
    10. △\triangle CAO (C, A, O)
    11. △\triangle DAO (D, A, O)
    12. △\triangle CBO (C, B, O)
    13. △\triangle DBO (D, B, O)
    14. △\triangle CAB (CAO + CBO)
    15. △\triangle DAB (DAO + DBO)
    16. △\triangle BHJ
    17. △\triangle BIJ
    18. △\triangle BHI (BHJ + BIJ)
    19. △\triangle ACO (No, O is on CD)
    20. △\triangle ADO (No)
    21. △\triangle ACD (No)
    22. △\triangle BCD (No)

    Let's use a simpler method.
    Count triangles in the top half (above CD):

    1. △\triangle AGE
    2. △\triangle AGF
    3. △\triangle AEF (1+2)
    4. △\triangle CGO
    5. △\triangle DGO
    6. △\triangle CAO (C, A, O)
    7. △\triangle DAO (D, A, O)
    8. △\triangle ACD (CAO + DAO) Total = 8 triangles in the top half.

    Count triangles in the bottom half (below CD): 9. △\triangle BHJ 10. △\triangle BIJ 11. △\triangle BHI (9+10) 12. △\triangle CJO 13. △\triangle DJO 14. △\triangle CBO (C, B, O) 15. △\triangle DBO (D, B, O) 16. △\triangle BCD (CBO + DBO) Total = 8 triangles in the bottom half.

    Now, count triangles that span across the horizontal line CD.
    These are triangles with base on AB. 17. △\triangle CGJ (CGO + CJO) 18. △\triangle DGJ (DGO + DJO) 19. △\triangle CAB (CAO + CBO) 20. △\triangle DAB (DAO + DBO) Total = 4 triangles.

    Total = 8 (top) + 8 (bottom) + 4 (spanning) = 20 triangles.

    Let's verify this count.
    Top section (vertex A):

    1. △\triangle AGE
    2. △\triangle AGF
    3. △\triangle AEF (AGE + AGF)
    4. △\triangle ACO (C, A, O)
    5. △\triangle ADO (D, A, O)
    6. △\triangle ACD (ACO + ADO) Total = 6.

    Bottom section (vertex B): 7. △\triangle BHJ 8. △\triangle BIJ 9. △\triangle BHI (BHJ + BIJ) 10. △\triangle BCO (C, B, O) 11. △\triangle BDO (D, B, O) 12. △\triangle BCD (BCO + BDO) Total = 6.

    Middle section (base on AB): 13. △\triangle CGO 14. △\triangle DGO 15. △\triangle CJO 16. △\triangle DJO 17. △\triangle CGJ (CGO + CJO) 18. △\triangle DGJ (DGO + DJO) 19. △\triangle CAO (C, A, O) 20. △\triangle DAO (D, A, O) 21. △\triangle CBO (C, B, O) 22. △\triangle DBO (D, B, O) 23. △\triangle CAB (CAO + CBO) 24. △\triangle DAB (DAO + DBO)

    Let's list all unique triangles:

    1. △\triangle AGE
    2. △\triangle AGF
    3. △\triangle AEF
    4. △\triangle CGO
    5. △\triangle DGO
    6. △\triangle CAO
    7. △\triangle DAO
    8. △\triangle ACD
    9. △\triangle BHJ
    10. △\triangle BIJ
    11. △\triangle BHI
    12. △\triangle CJO
    13. △\triangle DJO
    14. △\triangle CBO
    15. △\triangle DBO
    16. △\triangle BCD
    17. △\triangle CGJ (CGO + CJO)
    18. △\triangle DGJ (DGO + DJO)
    19. △\triangle CAB (CAO + CBO)
    20. △\triangle DAB (DAO + DBO)

    Total = 20 triangles.

    The final answer is 20\text{20}.

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