Difference between revisions of "2003 Indonesia MO Problems/Problem 2"
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Given a quadrilateral <math>ABCD</math>. Let <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are the midpoints of <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively. <math>PR</math> and <math>QS</math> intersects at <math>O</math>. Prove that <math>PO = OR</math> and <math>QO = OS</math>. | Given a quadrilateral <math>ABCD</math>. Let <math>P</math>, <math>Q</math>, <math>R</math>, and <math>S</math> are the midpoints of <math>AB</math>, <math>BC</math>, <math>CD</math>, and <math>DA</math>, respectively. <math>PR</math> and <math>QS</math> intersects at <math>O</math>. Prove that <math>PO = OR</math> and <math>QO = OS</math>. | ||
− | ==Solution== | + | ==Solution 1== |
<asy> | <asy> | ||
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Since <math>PQRS</math> is a parallelogram, <math>RS = QP.</math> In addition, by the Alternating Interior Angle Theorem, <math>\angle SRP = \angle RPQ</math> and <math>\angle RSQ = \angle SQP.</math> Thus, by ASA Congruency, <math>\triangle SRO \cong \triangle QPO.</math> Finally, using CPCTC shows that <math>RO = OP</math> and <math>SO = OQ.</math> | Since <math>PQRS</math> is a parallelogram, <math>RS = QP.</math> In addition, by the Alternating Interior Angle Theorem, <math>\angle SRP = \angle RPQ</math> and <math>\angle RSQ = \angle SQP.</math> Thus, by ASA Congruency, <math>\triangle SRO \cong \triangle QPO.</math> Finally, using CPCTC shows that <math>RO = OP</math> and <math>SO = OQ.</math> | ||
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+ | ==Solution 2== | ||
+ | Let <math>A = (A_x, A_y)</math>, <math>B = (B_x, B_y)</math>, <math>C = (C_x, C_y)</math>, and <math>D = (D_x, D_y)</math>. Then, we have <math>P = (\dfrac{A_x + B_x}{2}, \dfrac{A_y + B_y}{2})</math>, <math>Q = (\dfrac{B_x + C_x}{2}, \dfrac{B_y + C_y}{2})</math>, <math>R = (\dfrac{C_x + D_x}{2}, \dfrac{C_y + D_y}{2})</math>, and <math>S = (\dfrac{D_x + A_x}{2}, \dfrac{D_y + A_y}{2})</math>. Note that any line that goes through two points <math>X</math> and <math>Y</math> also goes through their midpoint. So, the line through <math>P</math> and <math>R</math> also goes through <math>(\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})</math>. Similarly, any line through <math>Q</math> and <math>S</math> also goes through <math>(\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})</math>. That means that both <math>\overline{PR}</math> and <math>\overline{QS}</math> go through <math>(\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})</math>, and because two non-identical lines that intersect only intersect once, <math>O = (\dfrac{A_x + B_x + C_x + D_x}{4}, \dfrac{A_y + B_y + C_y + D_y}{4})</math>. Since <math>O</math> is the midpoint of both <math>\overline{PR}</math> and <math>\overline{QS}</math>, we have proved that <math>PO = OR</math> and <math>QO = OS</math>. <math>\blacksquare</math> ~Puck_0 | ||
==See Also== | ==See Also== |
Latest revision as of 16:52, 22 January 2024
Contents
Problem
Given a quadrilateral . Let , , , and are the midpoints of , , , and , respectively. and intersects at . Prove that and .
Solution 1
Draw lines and . By SAS Similarity, and That means and making a parallelogram.
Since is a parallelogram, In addition, by the Alternating Interior Angle Theorem, and Thus, by ASA Congruency, Finally, using CPCTC shows that and
Solution 2
Let , , , and . Then, we have , , , and . Note that any line that goes through two points and also goes through their midpoint. So, the line through and also goes through . Similarly, any line through and also goes through . That means that both and go through , and because two non-identical lines that intersect only intersect once, . Since is the midpoint of both and , we have proved that and . ~Puck_0
See Also
2003 Indonesia MO (Problems) | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 | Followed by Problem 3 |
All Indonesia MO Problems and Solutions |