Fun Geometry Problem with Solution #150
⨷Âì 1 ![]() ¡Ó˹´ãËé x < 35° ¨§¾ÔÊÙ¨¹ìÇèÒ x = 30° ¾ÔÊÙ¨¹ì 1 ![]() (1) ∠BCD = 50° - x áÅÐ ∠ADC = 70° - x (2) µèÍ AC ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè CP = DP ⇒ ∠BCP = 20° + x ∵ CP = DP ⇔ ∆CDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´ ⇔ ∠CDP = ∠DCP ⇔ ∠CDP = 70° ⇔ ∠CPD = 40° ÊѧࡵÇèÒ CP = DP áÅÐ ∠CPD = 2(∠CBD) ⇒ ¨Ø´ P à»ç¹ circumcenter ¢Í§ ∆BCD ⇔ BP = CP = DP ãËé BP = L ⇒ CP = L áÅÐ DP = L (3) ¡Ó˹´¨Ø´ Q ãµé AD ·Õè·ÓãËé AQ = L áÅÐ ∠DAQ = 20° + x ¨ÐàËç¹ÇèÒ ∆ADQ ≅ ∆BCP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AD = BC, ∠DAQ = ∠BCP, AQ = CP) ⇒ DQ = BP (= L) ⇔ DQ = AQ ⇔ ∆ADQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´ ⇔ ∠ADQ = ∠DAQ ⇔ ∠ADQ = 20° + x ⇔ ∠AQD = 140° - 2x (4) ∵ DP = DQ ⇔ ∆DPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D (= 160°) à»ç¹ÁØÁÂÍ´ ⇔ ∠DPQ = 10° áÅÐ ∠DQP = 10° ⇔ ∠APQ = 30° áÅÐ ∠AQP = 130° - 2x (> 60° à¾ÃÒÐ x < 35°) (5) ¡Ó˹´¨Ø´ R à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ Q ¼èÒ¹ AP ⇒ ∆APR ≅ ∆APQ ⇒ PR = PQ, AR = AQ = L áÅÐ ∠APR = ∠APQ = 30° ∵ PQ = PR áÅÐ ∠QPR = 60° ⇒ ∆PQR à»ç¹ ∆´éÒ¹à·èÒ ⇒ PQ = QR ¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠PQR = 60° ⇔ ∠AQR = 70° - 2x (6) ÊѧࡵÇèÒ ∆AQR ≅ ∆DPQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (AQ = DP, AR = DQ, QR = PQ) ⇒ ∠AQR = ∠DPQ ⇔ 70° - 2x = 10° ⇔ x = 30° Q.E.D. ¾ÔÊÙ¨¹ì 2 ![]() (1) ∠BCD = 50° - x (2) µèÍ AC ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè CP = DP ⇒ ∠BCP = 20° + x ∵ CP = DP ⇔ ∆CDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´ ⇔ ∠CDP = ∠DCP ⇔ ∠CDP = 70° ⇔ ∠CPD = 40° ÊѧࡵÇèÒ CP = DP áÅÐ ∠CPD = 2(∠CBD) ⇒ ¨Ø´ P à»ç¹ circumcenter ¢Í§ ∆BCD ⇔ BP = CP = DP ãËé BP = L ⇒ CP = L áÅÐ DP = L (3) ¡Ó˹´¨Ø´ Q ãµé AD ·Õè·ÓãËé DQ = L áÅÐ ∠ADQ = 20° + x ¨ÐàËç¹ÇèÒ ∆ADQ ≅ ∆BCP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AD = BC, ∠ADQ = ∠BCP, DQ = CP) ⇒ AQ = BP ⇔ AQ = L (4) ¡Ó˹´¨Ø´ R º¹ AP ·Õè·ÓãËé DR = L ⇔ DR = DP ⇔ ∆DPR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D à»ç¹ÁØÁÂÍ´ ⇔ ∠DRP = ∠DPR ⇔ ∠DRP = 40° ⇔ ∠ADR = 40° - x (5) ∵ DQ = DR áÅÐ ∠QDR = 60° ⇒ ∆DQR à»ç¹ ∆´éÒ¹à·èÒ ⇒ QR = L áÅÐ ∠DQR = 60° ∵ AQ = DQ = QR ⇔ ¨Ø´ Q à»ç¹ circumcenter ¢Í§ ∆ADR ⇒ ∠DAR = (∠DQR)/2 ⇔ x = 30° Q.E.D. ⨷Âì 2 ![]() ¡Ó˹´ãËé x > 35° ¨§¾ÔÊÙ¨¹ìÇèÒ x = 40° ¾ÔÊÙ¨¹ì ![]() (1) ∠BCD = 50° - x áÅÐ ∠ADC = 70° - x (2) µèÍ AC ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè CP = DP ⇒ ∠BCP = 20° + x ∵ CP = DP ⇔ ∆CDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´ ⇔ ∠CDP = ∠DCP ⇔ ∠CDP = 70° ⇔ ∠CPD = 40° ÊѧࡵÇèÒ CP = DP áÅÐ ∠CPD = 2(∠CBD) ⇒ ¨Ø´ P à»ç¹ circumcenter ¢Í§ ∆BCD ⇔ BP = CP = DP ãËé BP = L ⇒ CP = L áÅÐ DP = L (3) ¡Ó˹´¨Ø´ Q ãµé AD ·Õè·ÓãËé AQ = L áÅÐ ∠DAQ = 20° + x ¨ÐàËç¹ÇèÒ ∆ADQ ≅ ∆BCP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AD = BC, ∠DAQ = ∠BCP, AQ = CP) ⇒ DQ = BP (= L) ⇔ DQ = AQ ⇔ ∆ADQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´ ⇔ ∠ADQ = ∠DAQ ⇔ ∠ADQ = 20° + x ⇔ ∠AQD = 140° - 2x (4) ∵ DP = DQ ⇔ ∆DPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D (= 160°) à»ç¹ÁØÁÂÍ´ ⇔ ∠DPQ = 10° áÅÐ ∠DQP = 10° ⇔ ∠APQ = 30° áÅÐ ∠AQP = 130° - 2x (< 60° à¾ÃÒÐ x > 35°) (5) ¡Ó˹´¨Ø´ R à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ Q ¼èÒ¹ AP ⇒ ∆APR ≅ ∆APQ ⇒ PR = PQ, AR = AQ = L áÅÐ ∠APR = ∠APQ = 30° ∵ PQ = PR áÅÐ ∠QPR = 60° ⇒ ∆PQR à»ç¹ ∆´éÒ¹à·èÒ ⇒ PQ = QR ¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠PQR = 60° ⇔ ∠AQR = 2x - 70° (6) ÊѧࡵÇèÒ ∆AQR ≅ ∆DPQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-´-´ (AQ = DP, AR = DQ, QR = PQ) ⇒ ∠AQR = ∠DPQ ⇔ 2x - 70° = 10° ⇔ x = 40° Q.E.D. ´Ù⨷Âì·Ñé§ËÁ´ Click !! Fun Geometry Problem with Solution #149
⨷Âì ![]() ¨§¾ÔÊÙ¨¹ìÇèÒ x = 10° ¾ÔÊÙ¨¹ì 1 ![]() ãËé AD = L (1) ∠ACB = 100° áÅÐ ∠ADB = 40° (2) ¡Ó˹´¨Ø´ P à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ D ¼èÒ¹ AB ⇒ ∆ABP ≅ ∆ABD ⇒ BP = BD, ∠ABP = ∠ABD = 30° áÅÐ ∠APB = ∠ADB = 40° ∵ BD = BP áÅÐ ∠DBP = 60° ⇒ ∆BDP à»ç¹ ∆´éÒ¹à·èÒ ⇒ BD = DP ¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠BDP = ∠BPD = 60° ⇔ ∠ADP = ∠APD = 20° (3) ¡Ó˹´¨Ø´ Q º¹ AC ·Õè·ÓãËé DQ = L ⇔ DQ = AD ⇔ ∆ADQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠D à»ç¹ÁØÁÂÍ´ ⇔ ∠AQD = ∠DAQ ⇔ ∠AQD = 80° ⇔ ∠ADQ = 20° ⇔ ∠BDQ = 20° (4) ÊѧࡵÇèÒ ∆BDQ ≅ ∆ADP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BD = DP, ∠BDQ = ∠ADP, DQ = AD) ⇒ ∠DBQ = ∠APD ⇔ ∠DBQ = 20° ⇔ ∠DBQ = ∠BDQ ⇔ ∆BDQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´ ⇔ BQ = DQ ⇔ BQ = L ¾Ô¨ÒÃ³Ò ∆BCQ ¨Ðä´éÇèÒ ∠BQC = 40° (5) ¾Ô¨ÒÃ³Ò ∆ABC ¨ÐàËç¹ÇèÒ ∠A = 30°, ∠C = 100° áÅÐÁըش Q º¹ AC ·Õè·ÓãËé ∠BQC = 40° ⇒ AC = BQ (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ìã¹â¨·Âì 1-1) ⇔ AC = L ⇔ AC = AD ⇔ ∆ACD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A (= 80°) à»ç¹ÁØÁÂÍ´ ⇔ ∠ADC (= ∠ACD) = 50° ⇔ ∠BDC = x = 10° Q.E.D. ¾ÔÊÙ¨¹ì 2 ![]() (1) ∠ACB = 100° áÅÐ ∠ADB = 40° (2) ¡Ó˹´¨Ø´ P à»ç¹¨Ø´µÑ´ÃÐËÇèÒ§ AC áÅÐ BD ∵ ∠BAP = ∠ABP ⇔ ∆ABP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´ ⇔ AP = BP (3) ¡Ó˹´¨Ø´ Q º¹ AD ·Õè·ÓãËé PQ = AP ⇔ ∆APQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠P à»ç¹ÁØÁÂÍ´ ⇔ ∠AQP = ∠PAQ ⇔ ∠AQP = 80° ⇒ ∠APQ = 20° áÅÐ ∠DPQ = 40° ∵ ∠PDQ = ∠DPQ ⇔ ∆DPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´ ⇔ DQ = PQ ⇔ DQ = AP (4) ¡Ó˹´¨Ø´ R º¹ PQ ·Õè·ÓãËé AR = AQ ⇔ ∆AQR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A à»ç¹ÁØÁÂÍ´ ⇔ ∠ARQ = ∠AQR ⇔ ∠ARQ = 80° ⇔ ∠ARP = 100° ÊѧࡵÇèÒ ∆BCP ≅ ∆APR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-Á-´ (∠BCP = ∠ARP, ∠CBP = ∠APR, BP = AP) ⇒ CP = AR ⇔ CP = AQ (5) ∵ AP = DQ áÅÐ CP = AQ ⇒ AP + CP = DQ + AQ ⇔ AC = AD ⇔ ∆ACD à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A (= 80°) à»ç¹ÁØÁÂÍ´ ⇔ ∠ADC (= ∠ACD) = 50° ⇔ ∠BDC = x = 10° Q.E.D. ´Ù⨷Âì·Ñé§ËÁ´ Click !! Fun Geometry Problem with Solution #148
⨷Âì ![]() ¨§¾ÔÊÙ¨¹ìÇèÒ x = 20° ¾ÔÊÙ¨¹ì 1 ![]() (1) ∠ACB = 80° áÅÐ ∠ADB = 50° (2) µèÍ CA ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè CP = BC ⇔ ∆BCP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C (= 80°) à»ç¹ÁØÁÂÍ´ ⇔ ∠CBP (= ∠BPC) = 50° ⇔ ∠ABP = 10° (3) ÊѧࡵÇèÒ ∆ABD ≅ ∆ABP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-Á-´ (∠ADB = ∠APB, ∠ABD = ∠ABP, AB = AB) ⇒ BD = BP ⇔ ∆BDP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B (= 20°) à»ç¹ÁØÁÂÍ´ ⇔ ∠BPD (= ∠BDP) = 80° ⇔ ∠CPD = 30° ⇔ ∠CPD = ∠CBD ⇔ ☐BCDP ÊÒÁÒöṺã¹Ç§¡ÅÁä´é ⇔ ∠DCP = ∠DBP ⇔ x = 20° Q.E.D. ¾ÔÊÙ¨¹ì 2 (äÁèãªé·Äɮպ·à¡ÕèÂǡѺǧ¡ÅÁ) ![]() (1) ∠ACB = 80° áÅÐ ∠ADB = 50° (2) µèÍ CA ÍÍ¡ä»Âѧ¨Ø´ P â´Â·Õè CP = BC ⇔ ∆BCP à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠C (= 80°) à»ç¹ÁØÁÂÍ´ ⇔ ∠CBP (= ∠BPC) = 50° ⇔ ∠ABP = 10° (3) ÊѧࡵÇèÒ ∆ABD ≅ ∆ABP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-Á-´ (∠ADB = ∠APB, ∠ABD = ∠ABP, AB = AB) ⇒ BD = BP (4) µèÍ PC ÍÍ¡ä»Âѧ¨Ø´ Q â´Â·Õè BQ = BP (= BD) ⇔ ∆BPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´ ⇔ ∠BQP = ∠BPQ ⇔ ∠BQP = 50° ⇔ ∠CBQ = 30° (5) ÊѧࡵÇèÒ ∆BCD ≅ ∆BCQ ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (BC = BC, ∠CBD = ∠CBQ, BD = BQ) ⇒ ∠BDC = ∠BQC ⇔ ∠BDC = 50° ⇔ ∠BCD = 100° ⇔ ∠ACD = x = 20° Q.E.D. ¾ÔÊÙ¨¹ì 3 (1) ∠ADB = 50° (2) ãËé α = 50° ¾Ô¨ÒÃ³Ò ☐ABCD ¨ÐàËç¹ÇèÒ ∠BAC = 60°, ∠CAD = 60°, ∠CBD = 30° áÅÐ ∠ADB = α ⇒ ∠BDC = α (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì) ⇔ ∠BDC = 50° (3) ¾Ô¨ÒÃ³Ò ∆ACD ¨Ðä´éÇèÒ ∠ACD = x = 20° Q.E.D. ´Ù⨷Âì·Ñé§ËÁ´ Click !! Fun Geometry Problem with Solution #147
⨷Âì ![]() ¨§¾ÔÊÙ¨¹ìÇèÒ x = 10° ¾ÔÊÙ¨¹ì 1 ![]() (1) ¾Ô¨ÒÃ³Ò ☐ACBP ¨Ðä´éÇèÒ ∠APB (ÁØÁãËè) = 360° - 150° ⇔ ∠APB (ÁØÁàÅç¡) = 150° (2) ¡Ó˹´¨Ø´ Q º¹ BC ·Õè·ÓãËé CQ = AC ¨ÐàËç¹ÇèÒ ∆CPQ ≅ ∆ACP ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (CP = CP, ∠PCQ = ∠ACP, CQ = AC) ⇒ PQ = AP ¹Í¡¨Ò¡¹Ñé¹ Âѧä´éÇèÒ ∠CQP = ∠CAP ⇔ ∠CQP = 20° ⇔ ∠BPQ = 10° ⇔ ∠BPQ = ∠PBQ ⇔ ∆BPQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´ ⇔ BQ = PQ ⇔ BQ = AP (3) µèÍ PQ ÍÍ¡ä»Âѧ¨Ø´ R â´Â·Õè BR = BP ⇔ ∆BPR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´ ⇔ ∠BRP = ∠BPR ⇔ ∠BRP = 10° ⇔ ∠PBR = 160° ⇔ ∠QBR = 150° (4) ÊѧࡵÇèÒ ∆ABP ≅ ∆BQR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ ´-Á-´ (AP = BQ, ∠APB = ∠QBR, BP = BR) ⇒ ∠ABP = ∠BRQ ⇔ x = 10° Q.E.D. ¾ÔÊÙ¨¹ì 2 ãËé α = 20°, β = 10° áÅÐ γ = 60° ⇒ α + β + γ = 90° ¾Ô¨ÒÃ³Ò ∆ABC ¨ÐàËç¹ÇèÒ Áըش P à»ç¹¨Ø´ÀÒÂã¹·Õè·ÓãËé ∠CAP = α, ∠CBP = β, ∠ACP = γ áÅÐ ∠BCP = γ ⇒ ∠ABP = β (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì) ⇔ x = 10° Q.E.D. ´Ù⨷Âì·Ñé§ËÁ´ Click !! Fun Geometry Problem with Solution #146
⨷Âì ![]() ¨§¾ÔÊÙ¨¹ìÇèÒ x = 10° ¾ÔÊÙ¨¹ì 1 ![]() (1) ∠ACB = 80° (2) ¡Ó˹´¨Ø´ O à»ç¹ circumcenter ¢Í§ ∆ABP ⇒ ... • AO = BO = OP • ∠AOP = 2(∠ABP) ⇔ ∠AOP = 80° • ∠BOP = 2(∠BAP) ⇔ ∠BOP = 20° ∵ AO = BO ⇔ ∆ABO à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠O (= 100°) à»ç¹ÁØÁÂÍ´ ⇔ ∠BAO = ∠ABO = 40° ãËé BO = L ⇔ OP = L (3) µèÍ BP Í͡仾º AC ·Õè¨Ø´ Q ⇒ ∠BQC = 80° ÊѧࡵÇèÒ ∆ABQ ≅ ∆ABO ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠BAQ = ∠BAO, AB = AB, ∠ABQ = ∠ABO) ⇒ BQ = BO ⇔ BQ = L ∵ ∠BQC = ∠BCQ ⇔ ∆BCQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠B à»ç¹ÁØÁÂÍ´ ⇔ BC = BQ ⇔ BC = L (4) ãËé α = 10° ÊѧࡵÇèÒ ☐BCPO à»ç¹ÊÕèàËÅÕèÂÁàÇéÒ ·ÕèÁÕ BC = BO = OP, ∠CBO = 120° - 2α áÅÐ ∠BOP = 2α ⇒ ∠BCP = α (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ìã¹â¨·Âì 3) ⇔ x = 10° Q.E.D. ¾ÔÊÙ¨¹ì 2 ![]() (1) µèÍ BP Í͡仾º AC ·Õè¨Ø´ Q ⇒ ∠APQ = 50° áÅÐ ∠BQC = 80° ∵ ∠BAQ = ∠ABQ ⇔ ∆ABQ à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠Q à»ç¹ÁØÁÂÍ´ ⇔ AQ = BQ ãËé BP = a áÅÐ PQ = b ⇒ BQ = a + b ⇔ AQ = a + b (2) µèÍ BQ ÍÍ¡ä»Âѧ¨Ø´ R â´Â·Õè AR = a + b ⇒ ∠AQR = 80° ∵ AQ = AR ⇔ ∆AQR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠A à»ç¹ÁØÁÂÍ´ ⇔ ∠ARQ = ∠AQR ⇔ ∠ARQ = 80° ⇔ ∠QAR = 20° (3) ∵ ∠PAR = ∠APR ⇔ ∆APR à»ç¹ ∆˹éÒ¨ÑèÇ ·ÕèÁÕ ∠R à»ç¹ÁØÁÂÍ´ ⇔ PR = AR ⇔ PR = a + b ⇔ QR = a ¨ÐàËç¹ÇèÒ ∆BCQ ≅ ∆AQR ´éǤÇÒÁÊÑÁ¾Ñ¹¸ìẺ Á-´-Á (∠CBQ = ∠QAR, BQ = AQ, ∠BQC = ∠AQR) ⇒ CQ = QR ⇔ CQ = a (4) ¾Ô¨ÒÃ³Ò ∆BCQ ¨ÐàËç¹ÇèÒ ∠B = 20°, ∠Q = 80° áÅÐÁըش P º¹ BQ ·Õè·ÓãËé BP = CQ ⇒ ∠BCP = 10° (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì) ⇔ x = 10° Q.E.D. ´Ù⨷Âì·Ñé§ËÁ´ Click !! |
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