Fun Geometry Problem with Solution #150
⨷Âì 1



¡Ó˹´ãËé x < 35°
¨§¾ÔÊÙ¨¹ìÇèÒ x = 30°
¾ÔÊÙ¨¹ì 1



(1) ∠BCD = 50° - x áÅР∠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 áÅР∠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 áÅР∠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° áÅР∠DQP = 10°      ∠APQ = 30° áÅР∠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 áÅР∠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 áÅР∠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 áÅР∠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 áÅР∠QDR = 60°      ∆DQR à»ç¹ ∆´éÒ¹à·èÒ      QR = L áÅР∠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 áÅР∠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 áÅР∠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 áÅР∠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° áÅР∠DQP = 10°      ∠APQ = 30° áÅР∠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 áÅР∠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.

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Fun Geometry Problem with Solution #149
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 10°
¾ÔÊÙ¨¹ì 1



ãËé AD = L

(1) ∠ACB = 100° áÅР∠ADB = 40°

(2) ¡Ó˹´¨Ø´ P à»ç¹ÀÒ¾Êзé͹¢Í§¨Ø´ D ¼èÒ¹ AB      ∆ABP  ∆ABD      BP = BD, ∠ABP = ∠ABD = 30° áÅР∠APB = ∠ADB = 40°
∵ BD = BP áÅР∠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¨ÒóҠ∆BCQ ¨Ðä´éÇèÒ ∠BQC = 40°

(5) ¾Ô¨ÒóҠ∆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° áÅР∠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° áÅР∠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.

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Fun Geometry Problem with Solution #148
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 20°
¾ÔÊÙ¨¹ì 1



(1) ∠ACB = 80° áÅР∠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° áÅР∠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¨ÒóҠ☐ABCD ¨ÐàËç¹ÇèÒ ∠BAC = 60°, ∠CAD = 60°, ∠CBD = 30° áÅР∠ADB = α      ∠BDC = α (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì)      ∠BDC = 50°

(3) ¾Ô¨ÒóҠ∆ACD ¨Ðä´éÇèÒ ∠ACD = x = 20°   Q.E.D.

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Fun Geometry Problem with Solution #147
⨷Âì



¨§¾ÔÊÙ¨¹ìÇèÒ x = 10°
¾ÔÊÙ¨¹ì 1



(1) ¾Ô¨ÒóҠ☐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° áÅРγ = 60°   ⇒   α + β + γ = 90°
¾Ô¨ÒóҠ∆ABC ¨ÐàËç¹ÇèÒ Áըش P à»ç¹¨Ø´ÀÒÂã¹·Õè·ÓãËé ∠CAP = α, ∠CBP = β, ∠ACP = γ áÅР∠BCP = γ      ∠ABP = β (Click à¾×èÍ´ÙÇÔ¸Õ¾ÔÊÙ¨¹ì)      x = 10°   Q.E.D.

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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 áÅР∠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¨ÒóҠ∆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.

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