triangle, circle and tangents

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

triangle, circle and tangents

G H J

Reveal topic tags

BritishMathematicalOlympiad

L

G H BBookmark kLocked kLocked NReply

Source: BrMO 2016/17 round 1 question 5

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

12750 posts

#1 h Dec 3, 2016, 8:38 AM • 2 Y

Y by Adventure10, Mango247

Let $ABC$ be a triangle with $\angle A < \angle B < 90^{\circ}$ and let $\Gamma$ be the circle through $A, B$ and $C$ . The tangents to $\Gamma$ at $A$ and $C$ meet at $P$ . The line segments $AB$ and $PC$ produced meet at $Q$ . It is given that

$[ACP] = [ABC] = [BCQ]$ .

Prove that $\angle BCA = 90^{\circ}$ . Here $[XYZ]$ denotes the area of triangle $XYZ$ .

See here: https://bmos.ukmt.org.uk/home/bmo1-2017.pdf

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

95 posts

#2 h Dec 3, 2016, 8:42 AM • 3 Y

Y by AndrewTom, Adventure10, Mango247

$[asy] unitsize(2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pair B=dir(0), A=-B, Q=(3,0), C=tangent(Q,origin,1), P=2/(A+conj(C)); D(arc(origin,A,B,CW),red); D(A--P--Q--A--C--B); D("A",A,SW); D("B",B,SE); D("C",C,dir(C)); D("P",P,dir(P)); D("Q",Q,SE); [/asy]$
First, $[ABC]=[BQC]\iff BQ=AB$ , so $\frac{b^2}{a^2}=\frac{QC^2}{QB^2}=\frac{QA}{QB}=2$ .

Now $[ACP]=[ABC]\implies b\cdot AP=ca\implies\frac{b^2}{2\cos B}=ca\implies\cos B=\frac{b^2}{2ca}=\frac{a}{c}$ , so $\angle BCA=90^{\circ}$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

12750 posts

#3 h Dec 3, 2016, 9:51 AM • 1 Y

Y by Adventure10

Very nice, liberator!

Are there any other ways of doing this?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

176 posts

#4 h Dec 3, 2016, 12:36 PM • 2 Y

Y by Adventure10, Mango247

liberator wrote:

$[asy] unitsize(2cm); pointpen=black; pathpen=rgb(0.4,0.6,0.8); pointfontpen=fontsize(10); pair B=dir(0), A=-B, Q=(3,0), C=tangent(Q,origin,1), P=2/(A+conj(C)); D(arc(origin,A,B,CW),red); D(A--P--Q--A--C--B); D("A",A,SW); D("B",B,SE); D("C",C,dir(C)); D("P",P,dir(P)); D("Q",Q,SE); [/asy]$
First, $[ABC]=[BQC]\iff BQ=AB$ , so $\frac{b^2}{a^2}=\frac{QC^2}{QB^2}=\frac{QA}{QB}=2$ .

Now $[ACP]=[ABC]\implies b\cdot AP=ca\implies\frac{b^2}{2\cos B}=ca\implies\cos B=\frac{b^2}{2ca}=\frac{a}{c}$ , so $\angle BCA=90^{\circ}$ .

Can you please explain me where did you get the $b^2/a^2$ part from?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

12750 posts

#5 h Dec 4, 2016, 10:50 AM • 2 Y

Y by Adventure10, Mango247

Hi ninjalearner.

I'll try to fill in some of the details in liberator's proof:

$[ABC] = [BCQ] \implies BQ = AB$ , since these triangles have the same perpendicular height with respect to the base $ABQ$ .

Now $QC$ is a tangent, so, by the alternate angle theorem, $\angle QCB = \angle CAB = \alpha$ , say.

$PA = PC$ because $PA$ and $PC$ are common tangents from the point $P$ and so their lengths are equal. This means that $\triangle PAC$ is isosceles, with $\angle CAP = \angle ACP = \beta$ , say. And, because $PC$ is a tangent, $\angle CBA = \beta$ , by the alternate segment theorem.

Since $ABQ$ is a straight line (segment), $\angle QBC = \pi - \beta$ , that is $\angle QBC = 180^{\circ} - \beta$ . So, the sine rule in triangles $ABC$ and $BQC$ give

$\frac{b}{\sin \beta} =\frac{a}{\sin \alpha}$ and $\frac{QC}{\sin (180^{\circ} - \beta)}=\frac{QB} {\sin \alpha}$ .

But $\sin (180^{\circ} - \beta) = \sin \beta$ , so

$\frac{b}{a} = \frac{QC}{QB}$

and therefore $\frac{b^{2}}{a^{2}} = \frac{QC^{2}}{QB^{2}}$ (*)

Now, by the tangent-secant theorem, $QC^{2} = QA \cdot QB = 2 QB \cdot QB = 2 QB^{2}$ , since $QB = AB \implies QA = 2QB$ , and so (*) becomes $\frac{b^{2}}{a^{2}} = 2$ .

Next, for liberator's second line:

$\angle PCA = \angle CBA = \beta$ , by the alternate angle theorem and $\angle PAC = \beta$ because triangle $PAC$ is isosceles.

Therefore, since $[ACP] = [ABC]$ , we have $\frac{1}{2} AP \cdot b \sin \beta = \frac{1}{2} a c \sin \beta$ ,

so $b \cdot AP = ac$ (**)

Now, if we bisect $AC$ at $N$ , say, then, from $\triangle APN$ (or, $\triangle PNC$ ), $\cos \beta = \frac{\frac{b}{2}}{AP}$ , giving $AP = \frac{b}{2 \cos \beta}$ .

Thus (**) becomes $\frac{b^{2}}{2 \cos \beta} = ac$ , so that

$\cos \beta = \frac{b^{2}}{2ac} = \frac{2a^{2}}{2ac} = \frac{a}{c}$ , because $\frac{b^{2}}{c^{2}} = 2$ , from the line after (*),

and so, $\angle BCA = 90^{\circ}$ .

P.S. Am I correct in stating that, because $\cos \beta = \frac{a}{c}$ , $\angle BCA$ must equal $90^{\circ}$ or does something else need to be said so that it must be true?

This post has been edited 3 times. Last edited by AndrewTom, Dec 4, 2016, 6:55 PM
Reason: Corrected a few typos. Hope it's ok now.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

12750 posts

#6 h Dec 10, 2016, 10:51 PM • 2 Y

Y by Adventure10, Mango247

Are there any other ways of doing this?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

4894 posts

#7 h Dec 10, 2016, 11:39 PM • 2 Y

Y by AndrewTom, Adventure10

Let $AB=c$
$[ABC]=[BQC]\to BQ=AB=c$
$CQ^2=AB*AQ=2c^2 \to CQ=c\sqrt{2}$

$AP=PC$
$AP*AC=AB*CB$ and $AC*AB=CB*CQ$
$AB=\frac{AP*AC}{CB}=\frac{CQ*AP}{AB}$
$AP=\frac{AB^2}{CQ}=\frac{c}{\sqrt{2}}$
$AP^2+AQ^2-PQ^2=AP^2+4AB^2-PC^2-CQ^2-2*PC*CQ=4c^2-2c^2-2c^2=0$ , so $\angle PAQ=90^{\circ}=\angle CAB+\angle CBA=180^{\circ}-\angle ACB \to \angle ACB=90^{\circ}$

This post has been edited 1 time. Last edited by RagvaloD, Dec 10, 2016, 11:41 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

7936 posts

#8 h Dec 10, 2016, 11:44 PM • 3 Y

Y by AndrewTom, Adventure10, Mango247

Here's a short solution that does not use any trigonometry.

Use Liberator's diagram above. Note that $[ABC]=[BCQ]$ implies that $AB=BQ$ . Thus, $2d(B,PQ) = d(A,PQ)$ (where here $d(A,\ell)$ denotes the distance from $A$ to $\ell$ ), and from $[APC]=[BCQ]$ we obtain $CQ=2PC=2PA\implies PQ=3PA$ .

Now let the angle bisector of $\angle APQ$ intersect $AQ$ at a point $X$ . Then by the Angle Bisector Theorem $\dfrac{AX}{XQ}=\dfrac{AP}{PQ}=\frac13\quad\implies\quad AX=\frac14AQ=\frac12AB,$ so in fact $X$ is the midpoint of $\overline{AB}$ . But since $X$ lies on the angle bisector of $\angle APC$ , we obtain $AX=XC$ . Thus $X$ is the circumcenter of $\triangle ABC$ , and since $X$ lies on $AB$ , we obtain the desired conclusion. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

12750 posts

#9 h Dec 11, 2016, 8:39 PM • 1 Y

Y by Adventure10

Thanks RagvaloD and djmathman.

Has anyone got another proof?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

241 posts

#10 h Dec 14, 2016, 9:09 PM • 3 Y

Y by AndrewTom, Adventure10, Mango247

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a