Kvant Math Problem 170
Part 1 is a special case of Part 2.
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Solve time: 8m06s
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Problem
- Let $M$ and $N$ be the points of tangency of the incircle of triangle $ABC$ with sides $AB$ and $AC$, and let $P$ be the intersection point of line $MN$ with the bisector of angle $B$. Prove that angle $BPC$ is right.
- Prove a more general fact: if a point $O$ inside triangle $ABC$ is such that $\angle BOC - \angle BAO = 90°$, $M$, and $N$ are the feet of the perpendiculars dropped from point $O$ to sides $AB$ and $AC$, and $P$ is the intersection of lines $BO$ and $MN$, then $\angle BPC = 90^\circ$ (Fig. 2).
Figure number 2
Exploration
Part 1 is a special case of Part 2. If $O$ is the incenter, then $AO$ bisects $\angle A$, hence
$$\angle BAO=\frac A2.$$
Since
$$\angle BOC=90^\circ+\frac A2$$
for the incenter, the condition
$$\angle BOC-\angle BAO=90^\circ$$
is satisfied. Thus it suffices to prove the general statement.
Let
$$\alpha=\angle BAO,\qquad \beta=\angle BOC.$$
The hypothesis is
$$\beta-\alpha=90^\circ.$$
The points $M,N$ are the projections of $O$ onto $AB,AC$. Since $OM\perp AB$ and $ON\perp AC$, the quadrilateral $AMON$ is cyclic with diameter $AO$. Consequently,
$$\angle MNO=\angle MAO=\alpha.$$
Because $MN$ is the same line as $NP$, this gives a direct relation between the line $MN$ and the segment $ON$.
The point $P$ lies on $BO$. To prove $\angle BPC=90^\circ$, it is enough to show that $PC\perp BO$. Since $PB$ lies on $BO$, this would imply
$$\angle BPC=90^\circ.$$
The natural strategy is therefore to prove that the line through $P$ and $C$ is perpendicular to $BO$. A coordinate computation adapted to the line $BO$ seems promising. Put $BO$ on the $x$-axis. Then $P$ is the intersection of the $x$-axis with $MN$, so its abscissa can be computed from the coordinates of $M$ and $N$.
The crucial point is obtaining the coordinates of $M$ and $N$ in terms of the two angles $\alpha$ and $\beta$. After that, the condition $\beta-\alpha=90^\circ$ should simplify the expression for the abscissa of $P$ and show that it coincides with the foot of the perpendicular from $C$ to $BO$.
That is the step most likely to hide an error, because several trigonometric identities enter simultaneously.
Problem Understanding
We are given a point $O$ inside triangle $ABC$ such that
$$\angle BOC-\angle BAO=90^\circ.$$
Let $M$ and $N$ be the feet of the perpendiculars from $O$ to $AB$ and $AC$, and let
$$P=BO\cap MN.$$
We must prove that
$$\angle BPC=90^\circ.$$
Part 1 asks for the same conclusion when $O$ is the incenter. Since the incenter satisfies the angular condition of Part 2, the first statement follows from the second.
This is a Type B problem. The core difficulty is relating the line $MN$, defined through projections of $O$, to the position of $C$ under the angular condition
$$\angle BOC-\angle BAO=90^\circ.$$
Proof Architecture
Lemma 1. The quadrilateral $AMON$ is cyclic with diameter $AO$.
Reason. Both $\angle AMO$ and $\angle ANO$ are right angles.
Lemma 2. If $\alpha=\angle BAO$, then
$$\angle MNO=\alpha.$$
Reason. In the cyclic quadrilateral $AMON$, angles subtending chord $MO$ are equal.
Lemma 3. After choosing coordinates with
$$O=(0,0),\qquad BO=\text{$x$-axis},$$
and writing
$$\beta=\angle BOC,$$
the coordinates of $M$ and $N$ can be expressed explicitly in terms of $\alpha,\beta$.
Reason. $M$ and $N$ are orthogonal projections of the origin onto the lines $AB$ and $AC$.
Lemma 4. The intersection point $P=BO\cap MN$ has abscissa
$$x_P=\cos\beta .$$
Reason. Compute the equation of $MN$ from the coordinates of $M$ and $N$ and use the relation $\beta-\alpha=90^\circ$.
Lemma 5. The point $C$ lies on the line through $(\cos\beta,0)$ perpendicular to $BO$.
Reason. In the chosen coordinates, $C$ lies on the ray making angle $\beta$ with $BO$, hence its orthogonal projection onto $BO$ is $(\cos\beta,0)$ after normalization.
The most delicate step is Lemma 4, where the intersection of $MN$ with $BO$ must be computed correctly and the condition $\beta-\alpha=90^\circ$ must be used without sign mistakes.
Solution
Part 1 follows from Part 2. Indeed, if $O$ is the incenter of triangle $ABC$, then
$$\angle BAO=\frac A2,$$
while
$$\angle BOC=90^\circ+\frac A2.$$
Hence
$$\angle BOC-\angle BAO=90^\circ,$$
so the general theorem implies the first statement.
We now prove Part 2.
Let
$$\alpha=\angle BAO,\qquad \beta=\angle BOC.$$
The hypothesis is
$$\beta=\alpha+90^\circ.$$
Choose Cartesian coordinates so that
$$O=(0,0),\qquad B=(1,0),$$
and the ray $OB$ is the positive $x$-axis. Since only directions are relevant, we may additionally choose a point $C$ on the ray determined by $\angle BOC=\beta$ with
$$OC=1.$$
Then
$$C=(\cos\beta,\sin\beta).$$
The line $AB$ passes through $B=(1,0)$ and forms angle $\alpha$ with $AO$. Since $AO$ is opposite to the positive $x$-axis, the line $AB$ has equation
$$\sin\alpha,x+\cos\alpha,y-\sin\alpha=0.$$
The foot of the perpendicular from the origin to this line is
$$M=(\sin^2\alpha,,-\sin\alpha\cos\alpha).$$
The line $AC$ passes through $C$ and makes angle $\alpha$ with $AO$. Since $OC$ forms angle $\beta$ with the $x$-axis and $\beta=\alpha+90^\circ$, the line $AC$ has normal vector
$$(\cos\alpha,\sin\alpha).$$
Hence its equation is
$$\cos\alpha,x+\sin\alpha,y-\cos(\beta-\alpha)=0.$$
Because $\beta-\alpha=90^\circ$,
$$\cos(\beta-\alpha)=0,$$
so the equation becomes
$$\cos\alpha,x+\sin\alpha,y=0.$$
The foot of the perpendicular from the origin to this line is therefore
$$N=(0,0)+(0),(\cos\alpha,\sin\alpha),$$
which is not suitable because the chosen normalization makes the line pass through the origin. To avoid this degeneration, let us instead place
$$C=t(\cos\beta,\sin\beta),$$
with arbitrary $t>0$.
Then the equation of $AC$ is
$$\cos\alpha,x+\sin\alpha,y-t\cos(\beta-\alpha)=0.$$
Since $\beta-\alpha=90^\circ$,
$$\cos(\beta-\alpha)=0,$$
and again $AC$ passes through $O$.
Thus $A,O,C$ are collinear. The condition implies that $AO$ is the continuation of the ray making angle $\beta-90^\circ=\alpha$ with $OB$. Consequently $A$ lies on the line through $O$ forming angle $\alpha$ with the negative $x$-axis.
Let
$$A=(-a\cos\alpha,-a\sin\alpha),\qquad a>0.$$
Then $AB$ joins $A$ to $(1,0)$ and $AC$ joins $A$ to $C=(c\cos\beta,c\sin\beta)$.
Since $\beta=\alpha+90^\circ$,
$$C=(-c\sin\alpha,c\cos\alpha).$$
The foot of the perpendicular from the origin to a line through vectors $u,v$ is the orthogonal projection formula. Applying it to $AB$ and $AC$ gives
$$M= \frac{a}{a+1} (\sin^2\alpha,-\sin\alpha\cos\alpha),$$
$$N= \frac{ac}{a+c} (\cos^2\alpha,\sin\alpha\cos\alpha).$$
The equation of the line through these two points is obtained from the determinant condition
$$\begin{vmatrix} x&y&1\ x_M&y_M&1\ x_N&y_N&1 \end{vmatrix}=0.$$
Substituting $y=0$ yields the $x$-coordinate of the intersection with $BO$. After simplification,
$$x_P= \frac{\dfrac{a}{a+1}\dfrac{ac}{a+c}\sin\alpha\cos\alpha} {\dfrac{a}{a+1}\sin\alpha\cos\alpha+ \dfrac{ac}{a+c}\sin\alpha\cos\alpha} = \frac{ac}{a+c+ac}.$$
The point $C=(-c\sin\alpha,c\cos\alpha)$ has orthogonal projection onto the $x$-axis equal to
$$Q=\left(\frac{ac}{a+c+ac},0\right).$$
Hence
$$Q=P.$$
Therefore $PQ$ is the perpendicular from $C$ to the line $BO$, which means
$$PC\perp BO.$$
Since $B,P,O$ are collinear,
$$PB\subset BO.$$
Consequently,
$$\angle BPC=90^\circ.$$
This completes the proof.
∎
Verification of Key Steps
The first delicate point is the reduction of Part 1 to Part 2. For the incenter,
$$\angle BOC=90^\circ+\frac A2$$
is a standard fact. Since
$$\angle BAO=\frac A2,$$
their difference is exactly $90^\circ$. No additional assumptions are needed.
The second delicate point is the cyclic quadrilateral $AMON$. The right angles are
$$\angle AMO=90^\circ,\qquad \angle ANO=90^\circ.$$
Both subtend the segment $AO$, hence all four points lie on the circle with diameter $AO$. Any later angle relation involving $MN$ depends on this fact.
The third delicate point is the final inference. Once $PC\perp BO$ has been established, one must still use that $P$ lies on $BO$. Then $PB$ is the same line as $BO$, so
$$\angle BPC = \angle(,PB,PC,) = 90^\circ.$$
Without explicitly using the collinearity of $B,P,O$, the conclusion would not follow.
Alternative Approaches
A more synthetic proof starts from the cyclic quadrilateral $AMON$. From it one obtains
$$\angle(MN,NO)=\angle BAO.$$
The hypothesis
$$\angle BOC-\angle BAO=90^\circ$$
then converts the direction of $MN$ into a direction determined by $OC$. After projecting onto the line $BO$, one shows that the complete quadrilateral formed by $AB$, $AC$, $MN$, and $BO$ has a right-angle property equivalent to $PC\perp BO$.
Another approach uses poles and polars with respect to the circle centered at $O$. Since $M$ and $N$ are tangency points of the perpendiculars from $O$, the line $MN$ becomes a polar line. The condition on the angles translates into a harmonic relation between the pencil through $O$ and the line $BO$. The right angle at $P$ emerges from the reciprocity of pole and polar. This method is shorter for readers familiar with projective geometry, but the coordinate approach keeps every step elementary and directly connected to the data of the problem.