Difference between revisions of "1996 IMO Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | {{ | + | <asy> |
+ | import graph; size(7cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -9.124923887131423, xmax = 11.886638474419073, ymin = -8.067061524000941, ymax = 7.112109861745523; /* image dimensions */ | ||
+ | pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); | ||
+ | /* draw figures */ | ||
+ | draw((-4.7910734582427414,-2.2520316878694198)--(-0.5209831486992185,-0.6439100533413341), linewidth(0.8) + wrwrwr); | ||
+ | draw((-1.5455209806493002,5.282227966879504)--(-0.5209831486992185,-0.6439100533413341), linewidth(0.8) + wrwrwr); | ||
+ | draw((-0.5209831486992185,-0.6439100533413341)--(4.6606059131635655,-2.3266536358088805), linewidth(0.8) + wrwrwr); | ||
+ | draw((-4.7910734582427414,-2.2520316878694198)--(-1.5455209806493002,5.282227966879504), linewidth(0.8) + wrwrwr); | ||
+ | draw((-1.5455209806493002,5.282227966879504)--(4.6606059131635655,-2.3266536358088805), linewidth(0.8) + wrwrwr); | ||
+ | draw((4.6606059131635655,-2.3266536358088805)--(-4.7910734582427414,-2.2520316878694198), linewidth(0.8) + wrwrwr); | ||
+ | draw((-2.663393291947277,2.6871874268422657)--(4.6606059131635655,-2.3266536358088805), linewidth(0.8) + linetype("4 4") + wrwrwr); | ||
+ | draw((0.4890818665379948,2.7877491336841196)--(-4.7910734582427414,-2.2520316878694198), linewidth(0.8) + linetype("4 4") + wrwrwr); | ||
+ | /* dots and labels */ | ||
+ | dot((-1.5455209806493002,5.282227966879504),linewidth(1pt) + dotstyle); | ||
+ | label("$A$", (-1.4792575117985467,5.317528023036582), NE * labelscalefactor); | ||
+ | dot((-4.7910734582427414,-2.2520316878694198),linewidth(1pt) + dotstyle); | ||
+ | label("$B$", (-5.368421189216425,-2.7776868388553828), NE * labelscalefactor); | ||
+ | dot((4.6606059131635655,-2.3266536358088805),linewidth(1pt) + dotstyle); | ||
+ | label("$C$", (4.727004680403201,-2.571154642954807), NE * labelscalefactor); | ||
+ | dot((-0.5209831486992185,-0.6439100533413341),linewidth(1pt) + dotstyle); | ||
+ | label("$P$", (-0.9754343965435009,-1.319589094904368), NE * labelscalefactor); | ||
+ | dot((-2.663393291947277,2.6871874268422657),linewidth(1pt) + dotstyle); | ||
+ | label("$F$", (-2.4513226744325554,2.6443488257930543), NE * labelscalefactor); | ||
+ | dot((0.4890818665379948,2.7877491336841196),linewidth(1pt) + dotstyle); | ||
+ | label("$G$", (0.6331148608484339,2.8125908731720175), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
+ | |||
+ | let <math>CF</math>, <math>BG</math> be the angle bisectors of <math>\angle ABP</math> and <math>\angle ACP</math>, respectively. Notice that they coincide with line <math>BD</math> and <math>CE</math>. Therefore, it suffices to show <math>CF,BG,AP</math> are concurrent. | ||
+ | Let <math>CF \cap AP = X_{1}</math>, and <math>BG\cap AP = X_{2}</math>. Notice that by angle bisector theorem, we have <cmath> | ||
+ | \frac{AX_{1}}{PX_{1}}=\frac{AC}{PC},\; | ||
+ | \frac{AX_{2}}{PX_{2}}=\frac{AB}{PB} | ||
+ | </cmath> | ||
+ | Therefore, it suffices to show <cmath>\frac{AC}{PC} =\frac{AB}{PB}</cmath> | ||
+ | |||
+ | <asy> | ||
+ | import graph; size(10cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -6.428011745667007, xmax = 17.921710402847665, ymin = -10.1071238327089, ymax = 7.546424724964209; /* image dimensions */ | ||
+ | pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); | ||
+ | /* draw figures */ | ||
+ | draw((-4.7910734582427414,-2.2520316878694198)--(-0.5209831486992185,-0.6439100533413341), linewidth(0.8) + wrwrwr); | ||
+ | draw((-1.5455209806493002,5.282227966879504)--(-0.5209831486992185,-0.6439100533413341), linewidth(0.8) + wrwrwr); | ||
+ | draw((-0.5209831486992185,-0.6439100533413341)--(4.6606059131635655,-2.3266536358088805), linewidth(0.8) + wrwrwr); | ||
+ | draw((-4.7910734582427414,-2.2520316878694198)--(-1.5455209806493002,5.282227966879504), linewidth(0.8) + wrwrwr); | ||
+ | draw((-1.5455209806493002,5.282227966879504)--(4.6606059131635655,-2.3266536358088805), linewidth(0.8) + wrwrwr); | ||
+ | draw((4.6606059131635655,-2.3266536358088805)--(-4.7910734582427414,-2.2520316878694198), linewidth(0.8) + wrwrwr); | ||
+ | draw((-1.5455209806493002,5.282227966879504)--(5.318779339995723,3.0814235558151237), linewidth(0.8) + wrwrwr); | ||
+ | draw((4.6606059131635655,-2.3266536358088805)--(5.318779339995723,3.0814235558151237), linewidth(0.8) + wrwrwr); | ||
+ | draw((-0.5209831486992185,-0.6439100533413341)--(5.318779339995723,3.0814235558151237), linewidth(0.8) + wrwrwr); | ||
+ | /* dots and labels */ | ||
+ | dot((-1.5455209806493002,5.282227966879504),linewidth(2pt) + dotstyle); | ||
+ | label("$A$", (-1.4493631992296316,5.328860743581626), NE * labelscalefactor); | ||
+ | dot((-4.7910734582427414,-2.2520316878694198),linewidth(2pt) + dotstyle); | ||
+ | label("$B$", (-5.210486701262846,-2.615204957788533), NE * labelscalefactor); | ||
+ | dot((4.6606059131635655,-2.3266536358088805),linewidth(2pt) + dotstyle); | ||
+ | label("$C$", (4.7467714546334765,-2.606539778032518), NE * labelscalefactor); | ||
+ | dot((-0.5209831486992185,-0.6439100533413341),linewidth(2pt) + dotstyle); | ||
+ | label("$P$", (-0.76232203188522677,-1.0411615600781519), NE * labelscalefactor); | ||
+ | dot((5.318779339995723,3.0814235558151237),linewidth(2pt) + dotstyle); | ||
+ | label("$Q$", (5.398996155040119,3.133037585545931), NE * labelscalefactor); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
+ | |||
+ | |||
+ | Now, construct <math>\triangle AQC \sim \triangle APB</math>. Connect <math>PQ</math>. We notice that <math>\angle BAC = \angle PAQ</math>, <math>\frac{AB}{AP}=\frac{AC}{AQ}</math>. Therefore <math>\triangle ABC \sim \triangle APQ</math>. Therefore, we have <cmath>\begin{align*} | ||
+ | \angle PQC &= \angle AQC -\angle AQP \ | ||
+ | &=\angle APB-\angle ACB \ | ||
+ | &=\angle APC-\angle ABC \ | ||
+ | &=\angle QPC | ||
+ | \end{align*}</cmath> | ||
+ | Therefore, <cmath>CP=CQ=\frac{AC}{AB} \cdot BP \Longrightarrow \frac{AC}{PC} =\frac{AB}{PB} \quad \blacksquare</cmath> | ||
+ | |||
+ | ~JoeyW | ||
==See Also== | ==See Also== |
Latest revision as of 13:46, 25 October 2024
Problem
Let be a point inside triangle such that
Let , be the incenters of triangles , , respectively. Show that , , meet at a point.
Solution
let , be the angle bisectors of and , respectively. Notice that they coincide with line and . Therefore, it suffices to show are concurrent. Let , and . Notice that by angle bisector theorem, we have Therefore, it suffices to show
Now, construct . Connect . We notice that , . Therefore . Therefore, we have
Therefore,
~JoeyW
See Also
1996 IMO (Problems) • Resources | ||
Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
All IMO Problems and Solutions |