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1996 IMO Problems/Problem 2

Revision as of 13:41, 25 October 2024 by Joeyw (talk | contribs)

Problem

Let $P$ be a point inside triangle $ABC$ such that

\[\angle APB-\angle ACB = \angle APC-\angle ABC\]

Let $D$, $E$ be the incenters of triangles $APB$, $APC$, respectively. Show that $AP$, $BD$, $CE$ meet at a point.

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 $CF$, $BG$ be the angle bisectors of $\angle ABP$ and $\angle ACP$, respectively. Notice that they coincide with line $BD$ and $CE$. Therefore, it suffices to show $CF,BG,AP$ are concurrent. Let $CF \cap AP = X_{1}$, and $BG\cap AP = X_{2}$. Notice that by angle bisector theorem, we have \[\frac{AX_{1}}{PX_{1}}=\frac{AC}{PC},\; \frac{AX_{2}}{PX_{2}}=\frac{AB}{PB}\] Therefore, it suffices to show \[\frac{AC}{PC} =\frac{AB}{PB}\]

[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 $\triangle AQC \sim \triangle APB$. Connect $PQ$. We notice that $\angle BAC = \angle PAQ$, $\frac{AB}{AP}=\frac{AC}{AQ}$. Therefore $\triangle ABC \sim \triangle APQ$. Therefore, we have \begin{align*} \angle PQC &= \angle AQC -\angle AQP \\ &=\angle APB-\angle ACB \\ &=\angle APC-\angle ABC \\ &=\angle QPC \end{align*} Therefore, \[CP=CQ=\frac{AC}{AB} \cdot BP \Longrightarrow \frac{AC}{PC} =\frac{AB}{PB} \quad \blacksquare\]

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
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